Problem: 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, …, p_n of the wise men, let's generate a binary string of length n-1: for each 1 ≤ 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 ≤ n ≤ 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 ≤ x < 2^{n-1}:

  * Let's consider a string s of length n-1, such that s_i = ⌊ \frac{x}{2^{i-1}} ⌋ mod 2 for all 1 ≤ i ≤ n - 1. 
  * The (x+1)-th number should be equal to the required answer for s. 

Examples

Input


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 (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct BigNum {
  vector<long long> value;
  void set(long long x) {
    value = *(new vector<long long>);
    value.push_back(x);
  }
  void duplicate(BigNum other) { value = other.value; }
  void add(BigNum other) {
    vector<long long> o = other.value;
    if (o.size() > value.size()) {
      vector<long long> t = value;
      value = o;
      o = t;
    }
    for (int i = 0; i < o.size(); i += 1) {
      value[i] += o[i];
      if (i < value.size() - 1 && value[i] >= (long long)1e9) {
        value[i + 1] += value[i] / (long long)1e9;
        value[i] = value[i] % (long long)1e9;
      }
    }
    if (value.back() >= (long long)1e9) {
      value.push_back(value.back() / (long long)1e9);
      value[value.size() - 2] = value[value.size() - 2] % (long long)1e9;
    }
  }
  void mul(long long mul) {
    for (int i = 0; i < value.size(); i += 1) {
      value[i] *= mul;
    }
    for (int i = 0; i < value.size() - 1; i += 1) {
      if (value[i] >= (long long)1e9) {
        value[i + 1] += value[i] / (long long)1e9;
        value[i] = value[i] % (long long)1e9;
      }
    }
    if (value.back() >= (long long)1e9) {
      value.push_back(value.back() / (long long)1e9);
      value[value.size() - 2] = value[value.size() - 2] % (long long)1e9;
    }
  }
  long double div(BigNum other1, BigNum other2) {
    int s = value.size();
    long double current = value.back();
    if (s >= 2) {
      current += ((long double)value[s - 2]) / (long long)1e9;
    }
    if (s >= 3) {
      current +=
          ((long double)value[s - 3]) / ((long long)1e9 * (long long)1e9);
    }
    int s1 = other1.value.size();
    int s2 = other2.value.size();
    long double other = 0;
    if (s1 + s2 - 1 == s) {
      other += other1.value.back() * other2.value.back();
      if (s1 >= 2) {
        other += ((long double)other1.value[s1 - 2] * other2.value.back()) /
                 (long long)1e9;
      }
      if (s2 >= 2) {
        other += ((long double)other2.value[s2 - 2] * other1.value.back()) /
                 (long long)1e9;
      }
      if (s1 >= 3) {
        other += ((long double)other1.value[s1 - 3] * other2.value.back()) /
                 ((long long)1e9 * (long long)1e9);
      }
      if (s2 >= 3) {
        other += ((long double)other2.value[s2 - 3] * other1.value.back()) /
                 ((long long)1e9 * (long long)1e9);
      }
    }
    if (s1 + s2 - 1 == s - 1) {
      other += other1.value.back() * other2.value.back();
      if (s1 >= 2) {
        other += ((long double)other1.value[s1 - 2] * other2.value.back()) /
                 (long long)1e9;
      }
      if (s2 >= 2) {
        other += ((long double)other2.value[s2 - 2] * other1.value.back()) /
                 (long long)1e9;
      }
      other /= (long long)1e9;
    }
    if (s1 + s2 - 1 == s - 2) {
      other += other1.value.back() * other2.value.back();
      other /= (long long)1e9 * (long long)1e9;
    }
    return other / current;
  }
  string stringify() {
    string out = to_string(value.back());
    for (int i = value.size() - 1 - 1; i > -1; i -= 1) {
      string cur = to_string(value[i] + (long long)1e9);
      out += cur.substr(1);
    }
    return out;
  }
};
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long triangle(long long x) { return ((x + 1) * x) / 2; }
long long antitriangle(long long x) {
  long long low = 0, high = 1000000007;
  while (low < high) {
    long long mid = (low + high) / 2;
    if (triangle(mid) < x) {
      low = mid + 1;
    } else {
      high = mid;
    }
  }
  return low;
}
long long modInverse(long long a) {
  long long m = 1000000007;
  long long y = 0, x = 1;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) {
    x += 1000000007;
  }
  return x;
}
long long modInverse(long long a, long long b) {
  long long m = b;
  long long y = 0, x = 1;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) {
    x += b;
  }
  return x;
}
long long pow(long long a, long long b) {
  if (a <= 1) {
    return a;
  }
  if (b == 0) {
    return 1;
  }
  if (b % 2 == 0) {
    return pow((a * a) % 1000000007, b / 2) % 1000000007;
  }
  return (a * pow((a * a) % 1000000007, b / 2)) % 1000000007;
}
vector<long long> dev(vector<pair<long long, long long> > divisors) {
  if (divisors.size() == 0) {
    vector<long long> cur;
    cur.push_back(1);
    return cur;
  }
  long long x = divisors.back().first;
  long long n = divisors.back().second;
  divisors.pop_back();
  vector<long long> ans = dev(divisors);
  vector<long long> cur;
  long long xi = 1;
  for (int i = 0; i < n + 1; i += 1) {
    for (int j = 0; j < ans.size(); j += 1) {
      cur.push_back(ans[j] * xi);
    }
    xi *= x;
  }
  return cur;
}
bool isprime(long long a) {
  if (a == 1) {
    return false;
  }
  for (int i = 2; i * i <= a; i++) {
    if (a % i == 0) {
      return false;
    }
  }
  return true;
}
void add(vector<int>& x, vector<int>& y) {
  for (int i = 0; i < x.size(); i += 1) {
    x[i] += y[i];
  }
}
vector<long long> mul(vector<long long> x, long long a) {
  for (int i = 0; i < x.size(); i += 1) {
    x[i] = (x[i] * a) % 1000000007;
  }
  return x;
}
vector<long long> add(vector<long long> x, vector<long long>& y) {
  for (int i = 0; i < x.size(); i += 1) {
    x[i] = (x[i] + y[i]) % 1000000007;
  }
  return x;
}
int lca(vector<vector<int> >& f, vector<int>& depth, int a, int b) {
  if (depth[a] < depth[b]) {
    swap(a, b);
  }
  for (int k = 0; k < 22; k += 1) {
    if ((depth[a] - depth[b]) & (1 << k)) {
      a = f[k][a];
    }
  }
  if (a == b) {
    return a;
  }
  for (int k = 22 - 1; k > -1; k -= 1) {
    if (f[k][a] != f[k][b]) {
      a = f[k][a];
      b = f[k][b];
    }
  }
  return f[0][a];
}
struct pt {
  long long x, y;
  long long d() const { return x * x + y * y; }
};
inline pt operator-(const pt& a) { return {-a.x, -a.y}; }
inline pt operator+(const pt& a, const pt& b) { return {a.x + b.x, a.y + b.y}; }
inline pt operator-(const pt& a, const pt& b) { return {a.x - b.x, a.y - b.y}; }
inline long long operator*(const pt& a, const pt& b) {
  return a.x * b.y - a.y * b.x;
}
inline bool operator<(const pt& a, const pt& b) { return a * b < 0; }
bool llll(const pt& a, const pt& b) { return a * b > 0; }
inline bool operator<=(const pt& a, const pt& b) { return a * b <= 0; }
inline long long operator/(const pt& a, const pt& b) {
  return a.x * b.x + a.y * b.y;
}
inline bool operator==(const pt& a, const pt& b) {
  return a.x == b.x && a.y == b.y;
}
inline bool operator!=(const pt& a, const pt& b) {
  return a.x != b.x || a.y != b.y;
}
pt o = {0, 0};
bool comp(pt a, pt b) { return (a - o) * (b - o) < 0; }
bool decreasing(int a, int b) { return a > b; }
struct booleantree {
  booleantree* left = NULL;
  booleantree* right = NULL;
  void add(int val) {
    booleantree* cur = this;
    for (int i = 30 - 1; i > -1; i -= 1) {
      if ((val & (1 << i))) {
        if (cur->right == NULL) {
          cur->right = new booleantree();
        }
        cur = cur->right;
      } else {
        if (cur->left == NULL) {
          cur->left = new booleantree();
        }
        cur = cur->left;
      }
    }
  }
  int find(int val) {
    booleantree* cur = this;
    int ans = 0;
    for (int i = 30 - 1; i > -1; i -= 1) {
      if ((val & (1 << i))) {
        if (cur->left == NULL) {
          cur = cur->right;
        } else {
          cur = cur->left;
          ans += (1 << i);
        }
      } else {
        if (cur->right == NULL) {
          cur = cur->left;
        } else {
          cur = cur->right;
          ans += (1 << i);
        }
      }
    }
    return ans;
  }
};
vector<int> dummy;
long long funcc(vector<int> x, int k) {
  vector<int> l;
  for (int i = 0; i < x.size(); i += 1) {
    if (x[i]) {
      l.push_back(i);
    }
  }
  long long ans = 0;
  vector<int> a(k);
  while (a[0] < l.size()) {
    int all = 0;
    for (int i = 0; i < k; i += 1) {
      all |= l[a[i]];
    }
    if (all == (1 << k) - 1) {
      long long sum = 1;
      int num = 1;
      for (int i = 0; i < k; i += 1) {
        sum *= x[l[a[i]]];
        x[l[a[i]]]--;
        if (i > 0 && a[i] == a[i - 1]) {
          num++;
          sum /= num;
        } else {
          num = 1;
        }
      }
      ans += sum;
      for (int i = 0; i < k; i += 1) {
        x[l[a[i]]]++;
      }
    }
    int i = k - 1;
    while (i > 0 && a[i] == l.size() - 1) {
      i--;
    }
    a[i++]++;
    while (i < k) {
      a[i] = a[i - 1];
      i++;
    }
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  vector<vector<int> > a(n, vector<int>(n));
  for (int i = 0; i < n; i += 1) {
    string l;
    cin >> l;
    for (int j = 0; j < n; j += 1) {
      a[i][j] = l[j] - '0';
    }
  }
  vector<vector<long long> > b(n * (1 << n));
  for (int i = 0; i < b.size(); i += 1) {
    long long lead = i % n;
    vector<int> in(n);
    long long count = 0;
    for (int j = 0; j < n; j += 1) {
      if ((i / n) & (1 << j)) {
        in[j] = 1;
        count++;
      }
    }
    if (!in[lead]) {
      continue;
    }
    long long rest = (i / n) - (1 << lead);
    b[i].resize(1 << (count - 1));
    if (count == 1) {
      b[i][0] = 1;
      continue;
    }
    for (int j = 0; j < n; j += 1) {
      if (in[j]) {
        int plus = a[lead][j] << (count - 2);
        for (int k = 0; k < b[j + rest * n].size(); k += 1) {
          b[i][plus + k] += b[j + rest * n][k];
        }
      }
    }
  }
  vector<long long> ans(1 << (n - 1));
  for (int i = n * ((1 << n) - 1); i < n * (1 << n); i += 1) {
    for (int j = 0; j < b[i].size(); j += 1) {
      ans[j] += b[i][j];
    }
  }
  for (int i = 0; i < 1 << (n - 1); i += 1) {
    cout << (ans[i]) << ' ';
  }
  cout << "" << '\n';
  return 0;
}