#include <bits/stdc++.h>
using namespace std;
int n, m;
long long f[1005][1005], g[1005][1005];
char s[1005][1005];
int main() {
  ios_base ::sync_with_stdio(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> (s[i] + 1);
    for (int j = 1; j <= m; j++) {
      f[i][j] = (s[i][j] == '*') + f[i][j - 1];
      g[i][j] = (s[i][j] == '*') + g[i - 1][j];
    }
  }
  long long res = 0LL;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (s[i][j] != '*') continue;
      res += (max(0LL, f[i][m] - 1) * max(0LL, g[n][j] - 1));
    }
  }
  cout << res << "\n";
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

You are given a n × m field consisting only of periods ('.') and asterisks ('*'). Your task is to count all right triangles with two sides parallel to the square sides, whose vertices are in the centers of '*'-cells. A right triangle is a triangle in which one angle is a right angle (that is, a 90 degree angle).

Input

The first line contains two positive integer numbers n and m (1 ≤ n, m ≤ 1000). The following n lines consist of m characters each, describing the field. Only '.' and '*' are allowed.

Output

Output a single number — total number of square triangles in the field. Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).

Examples

Input

2 2
**
*.


Output

1


Input

3 4
*..*
.**.
*.**


Output

9

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m;
long long f[1005][1005], g[1005][1005];
char s[1005][1005];
int main() {
  ios_base ::sync_with_stdio(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> (s[i] + 1);
    for (int j = 1; j <= m; j++) {
      f[i][j] = (s[i][j] == '*') + f[i][j - 1];
      g[i][j] = (s[i][j] == '*') + g[i - 1][j];
    }
  }
  long long res = 0LL;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (s[i][j] != '*') continue;
      res += (max(0LL, f[i][m] - 1) * max(0LL, g[n][j] - 1));
    }
  }
  cout << res << "\n";
  return 0;
}

```