#include <bits/stdc++.h>
using namespace std;
long long n, m, ans = 0;
long long dph[1002][1002], dpc[1002][1002];
int main() {
  cin >> n >> m;
  char c[n + 2][m + 2];
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cin >> c[i][j];
      if (c[i][j] == '*') {
        dph[i][j] = dph[i][j - 1] + 1;
        dpc[i][j] = dpc[i - 1][j] + 1;
      } else {
        dph[i][j] = dph[i][j - 1];
        dpc[i][j] = dpc[i - 1][j];
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (c[i][j] == '*') {
        int h = (dph[i][m] - 1) * (dpc[n][j] - 1);
        if (h > 0) ans = ans + h;
      }
    }
  }
  cout << ans;
}

### 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;
long long n, m, ans = 0;
long long dph[1002][1002], dpc[1002][1002];
int main() {
  cin >> n >> m;
  char c[n + 2][m + 2];
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cin >> c[i][j];
      if (c[i][j] == '*') {
        dph[i][j] = dph[i][j - 1] + 1;
        dpc[i][j] = dpc[i - 1][j] + 1;
      } else {
        dph[i][j] = dph[i][j - 1];
        dpc[i][j] = dpc[i - 1][j];
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (c[i][j] == '*') {
        int h = (dph[i][m] - 1) * (dpc[n][j] - 1);
        if (h > 0) ans = ans + h;
      }
    }
  }
  cout << ans;
}

```