#include <bits/stdc++.h>
int main() {
  int n, m;
  while (~scanf("%d %d", &n, &m)) {
    getchar();
    int l[1005] = {0}, u[1005] = {0};
    char mp[1005][1005];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        scanf("%c", &mp[i][j]);
        if (mp[i][j] == '*') {
          l[i]++;
          u[j]++;
        }
      }
      getchar();
    }
    long long sum = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        if (mp[i][j] == '*') {
          sum += (l[i] - 1) * (u[j] - 1);
        }
      }
    }
    printf("%lld\n", sum);
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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>
int main() {
  int n, m;
  while (~scanf("%d %d", &n, &m)) {
    getchar();
    int l[1005] = {0}, u[1005] = {0};
    char mp[1005][1005];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        scanf("%c", &mp[i][j]);
        if (mp[i][j] == '*') {
          l[i]++;
          u[j]++;
        }
      }
      getchar();
    }
    long long sum = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        if (mp[i][j] == '*') {
          sum += (l[i] - 1) * (u[j] - 1);
        }
      }
    }
    printf("%lld\n", sum);
  }
  return 0;
}

```