#include <bits/stdc++.h>
int main() {
  long long int n, m, i, j, k, count = 0, r[1001] = {0}, c[1001] = {0};
  scanf("%lld %lld", &n, &m);
  char a[1001][1001];
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      scanf(" %c", &a[i][j]);
      if (a[i][j] == '*') {
        r[i]++;
        c[j]++;
      }
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      if (a[i][j] == '*') {
        count = count + (r[i] - 1) * (c[j] - 1);
      }
    }
  }
  printf("%lld\n", count);
  return 0;
}

### 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>
int main() {
  long long int n, m, i, j, k, count = 0, r[1001] = {0}, c[1001] = {0};
  scanf("%lld %lld", &n, &m);
  char a[1001][1001];
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      scanf(" %c", &a[i][j]);
      if (a[i][j] == '*') {
        r[i]++;
        c[j]++;
      }
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      if (a[i][j] == '*') {
        count = count + (r[i] - 1) * (c[j] - 1);
      }
    }
  }
  printf("%lld\n", count);
  return 0;
}

```