# Wise Men (Easy Version)

This is the easy version of the problem. The difference is constraints on the number of wise men and the time limit. You can make hacks only if all versions of this task are solved.

$n$ wise men live in a beautiful city. Some of them know each other.

For each of the $n!$ possible permutations $p_1, p_2, \ldots, p_n$ of the wise men, let’s generate a binary string of length $n-1$: for each $1 \leq i < n$ set $s_i=1$ if $p_i$ and $p_{i+1}$ know each other, and $s_i=0$ otherwise.

For all possible $2^{n-1}$ binary strings, find the number of permutations that produce this binary string.Input

The first line of input contains one integer $n$ ($2 \leq n \leq 14)$  — the number of wise men in the city.

The next $n$ lines contain a binary string of length $n$ each, such that the $j$-th character of the $i$-th string is equal to ‘1’ if wise man $i$ knows wise man $j$, and equals ‘0’ otherwise.

It is guaranteed that if the $i$-th man knows the $j$-th man, then the $j$-th man knows $i$-th man and no man knows himself.Output

Print $2^{n-1}$ space-separated integers. For each $0 \leq x < 2^{n-1}$:

• Let’s consider a string $s$ of length $n-1$, such that $s_i = \lfloor \frac{x}{2^{i-1}} \rfloor \bmod 2$ for all $1 \leq i \leq n – 1$.
• The $(x+1)$-th number should be equal to the required answer for $s$.

Examplesinput

3
011
101
110


output

0 0 0 6


input

4
0101
1000
0001
1010


output

2 2 6 2 2 6 2 2


Note

In the first test, each wise man knows each other, so every permutation will produce the string $11$.

In the second test:

• If $p = \{1, 2, 3, 4\}$, the produced string is $101$, because wise men $1$ and $2$ know each other, $2$ and $3$ don’t know each other, and $3$ and $4$ know each other;
• If $p = \{4, 1, 2, 3\}$, the produced string is $110$, because wise men $1$ and $4$ know each other, $1$ and $2$ know each other and $2$, and $3$ don’t know each other;
• If $p = \{1, 3, 2, 4\}$, the produced string is $000$, because wise men $1$ and $3$ don’t know each other, $3$ and $2$ don’t know each other, and $2$ and $4$ don’t know each other.

Solution:

#undef _GLIBCXX_DEBUG

#include <bits/stdc++.h>

using namespace std;

int L[14][64];
int R[14][64];

int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
vector<string> s(n);
for (int i = 0; i < n; i++) {
cin >> s[i];
}
vector<string> old_s = s;
vector<long long> ret(1 << (n - 1));
int half = n / 2;
for (int sub = 0; sub < (1 << n); sub++) {
if (__builtin_popcount(sub) != half) {
continue;
}
vector<int> people;
for (int i = 0; i < n; i++) {
if (sub & (1 << i)) {
people.push_back(i);
}
}
for (int i = 0; i < n; i++) {
if (!(sub & (1 << i))) {
people.push_back(i);
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
s[i][j] = old_s[people[i]][people[j]];
}
}
memset(L, 0, sizeof(L));
memset(R, 0, sizeof(R));
{
vector<int> order(half);
iota(order.begin(), order.end(), 0);
do {
for (int i = 0; i < half - 1; i++) {
if (s[order[i]][order[i + 1]] == '1') {
}
}
} while (next_permutation(order.begin(), order.end()));
}
{
vector<int> order(n - half);
iota(order.begin(), order.end(), half);
do {
for (int i = 0; i < n - half - 1; i++) {
if (s[order[i]][order[i + 1]] == '1') {
}
}
} while (next_permutation(order.begin(), order.end()));
}
for (int whoR = half; whoR < n; whoR++) {
if (ft == 0) {
continue;
}
for (int who = 0; who < half; who++) {
int m = (maskR << half) | (s[whoR][who] == '1' ? (1 << (half - 1)) : 0);
int* LL = L[who];
}
}
}
}
}
for (int i = 0; i < (1 << (n - 1)); i++) {
if (i > 0) {
cout << " ";
}
cout << ret[i];
}
cout << '\n';
cerr << "time = " << clock() << " ms" << '\n';
return 0;
}