#include <bits/stdc++.h>
using namespace std;
long long n, m, r, c, ans, a[1010][1010], b[1010][1010];
string s;
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> s;
    for (int j = 1; j <= m; j++) {
      if (s[j - 1] == '*') a[i][j]++, b[i][j] = 1;
      a[i][j] += a[i - 1][j];
      a[i][j] += a[i][j - 1];
      a[i][j] -= a[i - 1][j - 1];
    }
  }
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (b[i][j]) {
        r = a[i - 1][j] - a[i - 1][j - 1] - a[0][j] + a[0][j - 1];
        c = a[i][m] - a[i][j] - a[i - 1][m] + a[i - 1][j];
        ans += r * c;
      }
    }
  }
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (b[i][j]) {
        r = a[i - 1][j] - a[i - 1][j - 1] - a[0][j] + a[0][j - 1];
        c = a[i][j - 1] - a[i][0] - a[i - 1][j - 1] + a[i - 1][0];
        ans += r * c;
      }
    }
  }
  for (int i = 1; i < n; i++) {
    for (int j = 1; j <= m; j++) {
      if (b[i][j]) {
        r = a[n][j] - a[n][j - 1] - a[i][j] + a[i][j - 1];
        c = a[i][m] - a[i][j] - a[i - 1][m] + a[i - 1][j];
        ans += r * c;
      }
    }
  }
  for (int i = 1; i < n; i++) {
    for (int j = 1; j <= m; j++) {
      if (b[i][j]) {
        r = a[n][j] - a[n][j - 1] - a[i][j] + a[i][j - 1];
        c = a[i][j - 1] - a[i][0] - a[i - 1][j - 1] + a[i - 1][0];
        ans += r * c;
      }
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve the following problem:

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, r, c, ans, a[1010][1010], b[1010][1010];
string s;
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> s;
    for (int j = 1; j <= m; j++) {
      if (s[j - 1] == '*') a[i][j]++, b[i][j] = 1;
      a[i][j] += a[i - 1][j];
      a[i][j] += a[i][j - 1];
      a[i][j] -= a[i - 1][j - 1];
    }
  }
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (b[i][j]) {
        r = a[i - 1][j] - a[i - 1][j - 1] - a[0][j] + a[0][j - 1];
        c = a[i][m] - a[i][j] - a[i - 1][m] + a[i - 1][j];
        ans += r * c;
      }
    }
  }
  for (int i = 2; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (b[i][j]) {
        r = a[i - 1][j] - a[i - 1][j - 1] - a[0][j] + a[0][j - 1];
        c = a[i][j - 1] - a[i][0] - a[i - 1][j - 1] + a[i - 1][0];
        ans += r * c;
      }
    }
  }
  for (int i = 1; i < n; i++) {
    for (int j = 1; j <= m; j++) {
      if (b[i][j]) {
        r = a[n][j] - a[n][j - 1] - a[i][j] + a[i][j - 1];
        c = a[i][m] - a[i][j] - a[i - 1][m] + a[i - 1][j];
        ans += r * c;
      }
    }
  }
  for (int i = 1; i < n; i++) {
    for (int j = 1; j <= m; j++) {
      if (b[i][j]) {
        r = a[n][j] - a[n][j - 1] - a[i][j] + a[i][j - 1];
        c = a[i][j - 1] - a[i][0] - a[i - 1][j - 1] + a[i - 1][0];
        ans += r * c;
      }
    }
  }
  cout << ans << endl;
  return 0;
}

```