A permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

Consider a permutation p of length n, we build a graph of size n using it as follows: 

  * For every 1 ≤ i ≤ n, find the largest j such that 1 ≤ j < i and p_j > p_i, and add an undirected edge between node i and node j 
  * For every 1 ≤ i ≤ n, find the smallest j such that i < j ≤ n and p_j > p_i, and add an undirected edge between node i and node j 



In cases where no such j exists, we make no edges. Also, note that we make edges between the corresponding indices, not the values at those indices.

For clarity, consider as an example n = 4, and p = [3,1,4,2]; here, the edges of the graph are (1,3),(2,1),(2,3),(4,3).

A permutation p is cyclic if the graph built using p has at least one simple cycle. 

Given n, find the number of cyclic permutations of length n. Since the number may be very large, output it modulo 10^9+7.

Please refer to the Notes section for the formal definition of a simple cycle

Input

The first and only line contains a single integer n (3 ≤ n ≤ 10^6).

Output

Output a single integer 0 ≤ x < 10^9+7, the number of cyclic permutations of length n modulo 10^9+7.

Examples

Input


4


Output


16

Input


583291


Output


135712853

Note

There are 16 cyclic permutations for n = 4. [4,2,1,3] is one such permutation, having a cycle of length four: 4 → 3 → 2 → 1 → 4.

Nodes v_1, v_2, …, v_k form a simple cycle if the following conditions hold: 

  * k ≥ 3. 
  * v_i ≠ v_j for any pair of indices i and j. (1 ≤ i < j ≤ k) 
  * v_i and v_{i+1} share an edge for all i (1 ≤ i < k), and v_1 and v_k share an edge. 
66
 652615
 482331
 336161
 33
 1000000
 79531
 768208
 3
 885131

257415584
 960319213
 722928541
 234634596
 762187807
 23581336
 162141608
 635322133
 2
 329995454

#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long INF = 5e18;
const long double pi = acos(-1.0);
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
int gcd(int a, int b) { return (b == 0 ? a : gcd(b, a % b)); }
int lcm(int a, int b) { return (a * b) / gcd(a, b); }
inline long long mod(long long n, long long m) {
  long long ret = n % m;
  if (ret < 0) ret += m;
  return ret;
}
long double rad(long double x) { return x * pi / 180.0; }
bool isleft(pair<int, int> a, pair<int, int> b, pair<int, int> c) {
  int det = (b.first - a.first) * (c.second - a.second) -
            (c.first - a.first) * (b.second - a.second);
  if (det >= 0) return true;
  if (det < 0) return false;
  return false;
}
long long exp(int a, int b) {
  int result = 1;
  while (b > 0) {
    if (b & 1) result = mod(result * a, MOD);
    b >>= 1;
    a = mod(a * a, MOD);
  }
  return result;
}
void solve() {
  long long n, m;
  cin >> n >> m;
  if (n > m) {
    swap(n, m);
  }
  int adj[n][m];
  int um = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      char c;
      cin >> c;
      if (c == '1') {
        adj[i][j] = 1;
      } else {
        adj[i][j] = 0;
      }
      um += adj[i][j];
    }
  }
  if (n >= 4 and m >= 4) {
    cout << -1 << "\n";
    return;
  }
  if (n == 1 or m == 1) {
    cout << 0 << "\n";
    return;
  }
  if (n == 2 and m == 2) {
    if (um & 1)
      cout << 0 << "\n";
    else
      cout << 1 << "\n";
    return;
  }
  int ans = 0;
  if (n == 2) {
    for (int k = 0; k < m - 2; k++) {
      int left = 0;
      int right = 0;
      for (int i = 0; i < 2; i++) {
        for (int j = k; j < k + 2; j++) {
          left += adj[i][j];
        }
      }
      for (int i = 0; i < 2; i++) {
        for (int j = k + 1; j < k + 3; j++) {
          right += adj[i][j];
        }
      }
      if (left % 2 == 0 and right % 2 == 0) {
        ans++;
        adj[0][k + 1]++;
        adj[0][k + 1] %= 2;
      }
      if (left & 1 and right % 2 == 0) {
        ans++;
        adj[0][k + 2]++;
        adj[0][k + 2] %= 2;
      }
      if (left % 2 == 0 and right & 1) {
        ans++;
        adj[0][k]++;
        adj[0][k] %= 2;
      }
    }
  }
  if (n == 3) {
    for (int l = 0; l <= 1; l++) {
      for (int k = 0; k < m - 2; k++) {
        int left = 0;
        int right = 0;
        for (int i = l; i < l + 2; i++) {
          for (int j = k; j < k + 2; j++) {
            left += adj[i][j];
          }
        }
        for (int i = l; i < l + 2; i++) {
          for (int j = k + 1; j < k + 3; j++) {
            right += adj[i][j];
          }
        }
        if (l == 0) {
          if (left % 2 == 0 and right % 2 == 0) {
            ans++;
            adj[0][k + 1]++;
            adj[0][k + 1] %= 2;
          }
          if (left & 1 and right % 2 == 0) {
            ans++;
            adj[0][k + 2]++;
            adj[0][k + 2] %= 2;
          }
          if (left % 2 == 0 and right & 1) {
            ans++;
            adj[0][k]++;
            adj[0][k] %= 2;
          }
        } else {
          if (left % 2 == 0 and right % 2 == 0) {
            ans++;
            adj[2][k + 1]++;
            adj[2][k + 1] %= 2;
          }
          if (left & 1 and right % 2 == 0) {
            ans++;
            adj[2][k + 2]++;
            adj[2][k + 2] %= 2;
          }
          if (left % 2 == 0 and right & 1) {
            ans++;
            adj[2][k]++;
            adj[2][k] %= 2;
          }
        }
      }
    }
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t = 1;
  while (t--) solve();
}
