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

### Prompt

Generate a CPP solution 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>
using namespace std;
char a[1005][1005];
long long b[1005];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      scanf(" %c", &a[i][j]);
      if (a[i][j] == '*') {
        ++b[j];
      }
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    long long sum = 0, tot = 0;
    for (int j = 1; j <= m; ++j) {
      if (a[i][j] == '*') {
        ++sum;
        tot += b[j] - 1;
      }
    }
    ans += (sum - 1) * tot;
  }
  printf("%lld\n", ans);
  return 0;
}

```