Desk Disorder

A new set of desks just arrived, and it’s about time! Things were getting quite cramped in the office. You’ve been put in charge of creating a new seating chart for the engineers. The desks are numbered, and you sent out a survey to the engineering team asking each engineer the number of the desk they currently sit at, and the number of the desk they would like to sit at (which may be the same as their current desk). Each engineer must either remain where they sit, or move to the desired seat they indicated in the survey. No two engineers currently sit at the same desk, nor may any two engineers sit at the same desk in the new seating arrangement.

How many seating arrangements can you create that meet the specified requirements? The answer may be very large, so compute it modulo 1000000007 = 10⁹ + 7.

Input

Input will begin with a line containing N (1 ≤ N ≤ 100000), the number of engineers.

N lines follow, each containing exactly two integers. The i-th line contains the number of the current desk of the i-th engineer and the number of the desk the i-th engineer wants to move to. Desks are numbered from 1 to 2·N. It is guaranteed that no two engineers sit at the same desk.

Output

Print the number of possible assignments, modulo 1000000007 = 10⁹ + 7.

Examples

input

4
1 5
5 2
3 7
7 3

output

6

input

5
1 10
2 10
3 10
4 10
5 5

output

5

Note

These are the possible assignments for the first example:

• 1 5 3 7
• 1 2 3 7
• 5 2 3 7
• 1 5 7 3
• 1 2 7 3
• 5 2 7 3

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

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

undigraph(int n) : graph<T>(n) {
}

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

class dsu {
  public:
  vector <int> p;
int n;

dsu(int n) : n(n) {
p.resize(n);
iota(p.begin(), p.end(), 0);
}

inline int get(int x) {
return (x == p[x] ? x : (p[x] = get(p[x])));
}

inline bool unite(int x, int y) {
x = get(x);
y = get(y);
if (x != y) {
p[x] = y;
return true;
}
return false;
}
};

template <typename T>
vector< vector<int> > find_cycles(const graph<T> &g, int bound = 1 << 30) {
  vector<int> was(g.n, -1);
vector<int> st;
vector< vector<int> > cycles;
function<void(int, int)> dfs = [&g, &was, &st, &cycles, &dfs, &bound](int v, int pe) {
if ((int) cycles.size() >= bound) {
return;
}
was[v] = st.size();
for (int id : g.g[v]) {
if (id == pe) {
continue;
}
auto &e = g.edges[id];
int to = e.from ^ e.to ^ v;
if (was[to] >= 0) {
vector<int> cycle(1, id);
for (int j = was[to]; j < (int) st.size(); j++) {
          cycle.push_back(st[j]);
        }
        cycles.push_back(cycle);
        if ((int) cycles.size() >= bound) {
return;
}
continue;
}
if (was[to] == -1) {
st.push_back(id);
dfs(to, id);
st.pop_back();
}
}
was[v] = -2;
};
for (int i = 0; i < g.n; i++) {
    if (was[i] == -1) {
      dfs(i, -1);
    }
  }
  return cycles;
  // returns at most bound cycles (given by edge ids)
  // digraph: finds at least one cycle in every connected component
  // undigraph: finds cycle basis
}

const int md = 1e9 + 7;

inline void add(int &a, int b) {
  a += b;
  if (a >= md) a -= md;
}

inline void sub(int &a, int b) {
a -= b;
if (a < 0) a += md;
}

inline int mul(int a, int b) {
#if !defined(_WIN32) || defined(_WIN64)
  return (long long) a * b % md;
#endif
  unsigned long long x = (long long) a * b;
  unsigned xh = (unsigned) (x >> 32), xl = (unsigned) x, d, m;
asm(
"divl %4; \n\t"
: "=a" (d), "=d" (m)
: "d" (xh), "a" (xl), "r" (md)
);
return m;
}

inline int power(int a, int b) {
int res = 1;
while (b > 0) {
if (b & 1) {
res = mul(res, a);
}
a = mul(a, a);
b >>= 1;
}
return res;
}

inline int inv(int a) {
a %= md;
if (a < 0) a += md;
  int b = md, u = 0, v = 1;
  while (a) {
    int t = b / a;
    b -= t * a; swap(a, b);
    u -= t * v; swap(u, v);
  }
  assert(b == 1);
  if (u < 0) u += md;
  return u;
}

int main() {
  int n;
  scanf("%d", &n);
  n *= 2;
  undigraph<int> g(n);
dsu d(n);
for (int i = 0; i < n / 2; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    x--; y--;
    g.add(x, y);
    d.unite(x, y);
  }
  vector<int> coeff(n, 0);
for (int i = 0; i < n; i++) {
    coeff[d.get(i)]++;
  }
  vector< vector<int> > cycles = find_cycles(g);
for (auto &cycle : cycles) {
auto &e = g.edges[cycle[0]];
coeff[d.get(e.from)] = min((int) cycle.size(), 2);
}
int ans = 1;
for (int i = 0; i < n; i++) {
    if (coeff[i] != 0) {
      ans = mul(ans, coeff[i]);
    }
  }
  printf("%d\n", ans);
  return 0;
}