Quantifier Question

Logical quantifiers are very useful tools for expressing claims about a set. For this problem, let’s focus on the set of real numbers specifically. The set of real numbers includes zero and negatives. There are two kinds of quantifiers: universal ($\forall$) and existential ($\exists$). You can read more about them here.

The universal quantifier is used to make a claim that a statement holds for all real numbers. For example:

• $\forall x,x<100$ is read as: for all real numbers $x$, $x$ is less than $100$. This statement is false.
• $\forall x,x>x-1$ is read as: for all real numbers $x$, $x$ is greater than $x-1$. This statement is true.

The existential quantifier is used to make a claim that there exists some real number for which the statement holds. For example:

• $\exists x,x<100$ is read as: there exists a real number $x$ such that $x$ is less than $100$. This statement is true.
• $\exists x,x>x-1$ is read as: there exists a real number $x$ such that $x$ is greater than $x-1$. This statement is true.

Moreover, these quantifiers can be nested. For example:

• $\forall x,\exists y,x<y$ is read as: for all real numbers $x$, there exists a real number $y$ such that $x$ is less than $y$. This statement is true since for every $x$, there exists $y=x+1$.
• $\exists y,\forall x,x<y$ is read as: there exists a real number $y$ such that for all real numbers $x$, $x$ is less than $y$. This statement is false because it claims that there is a maximum real number: a number $y$ larger than every $x$.

Note that the order of variables and quantifiers is important for the meaning and veracity of a statement.

There are $n$ variables $x_1,x_2,\ldots,x_n$, and you are given some formula of the form $$ f(x_1,\dots,x_n):=(x_{j_1}<x_{k_1})\land (x_{j_2}<x_{k_2})\land \cdots\land (x_{j_m}<x_{k_m}), $$

where $\land$ denotes logical AND. That is, $f(x_1,\ldots, x_n)$ is true if every inequality $x_{j_i}<x_{k_i}$ holds. Otherwise, if at least one inequality does not hold, then $f(x_1,\ldots,x_n)$ is false.

Your task is to assign quantifiers $Q_1,\ldots,Q_n$ to either universal ($\forall$) or existential ($\exists$) so that the statement $$ Q_1 x_1, Q_2 x_2, \ldots, Q_n x_n, f(x_1,\ldots, x_n) $$

is true, and the number of universal quantifiers is maximized, or determine that the statement is false for every possible assignment of quantifiers.

Note that the order the variables appear in the statement is fixed. For example, if $f(x_1,x_2):=(x_1<x_2)$ then you are not allowed to make $x_2$ appear first and use the statement $\forall x_2,\exists x_1, x_1<x_2$. If you assign $Q_1=\exists$ and $Q_2=\forall$, it will only be interpreted as $\exists x_1,\forall x_2,x_1<x_2$.Input

The first line contains two integers $n$ and $m$ ($2\le n\le 2\cdot 10^5$; $1\le m\le 2\cdot 10^5$) — the number of variables and the number of inequalities in the formula, respectively.

The next $m$ lines describe the formula. The $i$-th of these lines contains two integers $j_i$,$k_i$ ($1\le j_i,k_i\le n$, $j_i\ne k_i$).Output

If there is no assignment of quantifiers for which the statement is true, output a single integer $-1$.

Otherwise, on the first line output an integer, the maximum possible number of universal quantifiers.

On the next line, output a string of length $n$, where the $i$-th character is “A” if $Q_i$ should be a universal quantifier ($\forall$), or “E” if $Q_i$ should be an existential quantifier ($\exists$). All letters should be upper-case. If there are multiple solutions where the number of universal quantifiers is maximum, print any.Examplesinput

2 1
1 2

output

1
AE

input

4 3
1 2
2 3
3 1

output

-1

input

3 2
1 3
2 3

output

2
AAE

Note

For the first test, the statement $\forall x_1, \exists x_2, x_1<x_2$ is true. Answers of “EA” and “AA” give false statements. The answer “EE” gives a true statement, but the number of universal quantifiers in this string is less than in our answer.

For the second test, we can show that no assignment of quantifiers, for which the statement is true exists.

For the third test, the statement $\forall x_1, \forall x_2, \exists x_3, (x_1<x_3)\land (x_2<x_3)$ is true: We can set $x_3=\max\{x_1,x_2\}+1$.

Solution:

#include <bits/stdc++.h>

using namespace std;

template <typename T>
class graph {
public:
struct edge {
int from;
int to;
T cost;
};

vector<edge> edges;
vector<vector<int>> g;
int n;

graph(int _n) : n(_n) {
g.resize(n);
}

virtual int add(int from, int to, T cost) = 0;
};

template <typename T>
class digraph : public graph<T> {
public:
using graph<T>::edges;
using graph<T>::g;
using graph<T>::n;

digraph(int _n) : graph<T>(_n) {
}

int add(int from, int to, T cost = 1) {
assert(0 <= from && from < n && 0 <= to && to < n);
    int id = (int) edges.size();
    g[from].push_back(id);
    edges.push_back({from, to, cost});
    return id;
  }

  digraph<T> reverse() const {
digraph<T> rev(n);
for (auto &e : edges) {
rev.add(e.to, e.from, e.cost);
}
return rev;
}
};

template <typename T>
vector<int> find_topsort(const digraph<T> &g) {
vector<int> deg(g.n, 0);
for (int id = 0; id < (int) g.edges.size(); id++) {
    deg[g.edges[id].to]++;
  }
  vector<int> x;
for (int i = 0; i < g.n; i++) {
    if (deg[i] == 0) {
      x.push_back(i);
    }
  }
  for (int ptr = 0; ptr < (int) x.size(); ptr++) {
    int i = x[ptr];
    for (int id : g.g[i]) {
      auto &e = g.edges[id];
      int to = e.to;
      if (--deg[to] == 0) {
        x.push_back(to);
      }
    }
  }
  if ((int) x.size() != g.n) {
    return vector<int>();
}
return x;
}

int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n, m;
cin >> n >> m;
digraph<int> g(n);
for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
--x; --y;
g.add(x, y);
}
vector<int> order = find_topsort(g);
if (order.empty()) {
cout << -1 << '\n';
    return 0;
  }
  string res(n, 'A');
  vector<int> mn(n);
iota(mn.begin(), mn.end(), 0);
for (int i : order) {
for (int eid : g.g[i]) {
auto& e = g.edges[eid];
int j = e.to;
mn[j] = min(mn[j], mn[i]);
}
}
for (int i = 0; i < n; i++) {
    if (mn[i] < i) {
      res[i] = 'E';
    }
  }
  iota(mn.begin(), mn.end(), 0);
  for (int i : vector<int>(order.rbegin(), order.rend())) {
for (int eid : g.g[i]) {
auto& e = g.edges[eid];
int j = e.to;
mn[i] = min(mn[i], mn[j]);
}
}
for (int i = 0; i < n; i++) {
    if (mn[i] < i) {
      res[i] = 'E';
    }
  }
  int ans = 0;
  for (char c : res) {
    ans += (c == 'A');
  }
  cout << ans << '\n';
  cout << res << '\n';
  return 0;
}