#include <bits/stdc++.h>
using namespace std;
int a[1005], b[1005];
bool v[1005][1005];
char s[1005];
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%s", s);
    for (int j = 0; s[j]; j++) {
      if (s[j] == '*') {
        a[i]++;
        b[j]++;
        v[i][j] = 1;
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (!v[i][j]) continue;
      ans += (a[i] - 1) * (b[j] - 1);
    }
  }
  printf("%I64d\n", ans);
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 a[1005], b[1005];
bool v[1005][1005];
char s[1005];
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%s", s);
    for (int j = 0; s[j]; j++) {
      if (s[j] == '*') {
        a[i]++;
        b[j]++;
        v[i][j] = 1;
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (!v[i][j]) continue;
      ans += (a[i] - 1) * (b[j] - 1);
    }
  }
  printf("%I64d\n", ans);
}

```