#include <bits/stdc++.h>
int n, m, a[1010][2];
char c[1010][1010];
long long ans;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    getchar();
    for (int j = 1; j <= m; ++j) {
      c[i][j] = getchar();
      if (c[i][j] == '*') {
        ++a[i][0];
        ++a[j][1];
      }
    }
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j)
      if (c[i][j] == '*')
        ans = ans + ((long long)(a[i][0] - 1)) * ((long long)(a[j][1] - 1));
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Please create a solution in cpp to 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 n, m, a[1010][2];
char c[1010][1010];
long long ans;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    getchar();
    for (int j = 1; j <= m; ++j) {
      c[i][j] = getchar();
      if (c[i][j] == '*') {
        ++a[i][0];
        ++a[j][1];
      }
    }
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j)
      if (c[i][j] == '*')
        ans = ans + ((long long)(a[i][0] - 1)) * ((long long)(a[j][1] - 1));
  printf("%I64d\n", ans);
  return 0;
}

```