#include <bits/stdc++.h>
using namespace std;
long long hor[1001], ver[1001];
signed main() {
  long long n, m;
  cin >> n >> m;
  char a[n + 1][m + 1], z;
  long long i, j, x, y;
  for (i = 1; i <= n; ++i) {
    for (j = 1; j <= m; ++j) cin >> a[i][j];
  }
  for (i = 1; i <= n; ++i) {
    for (j = 1; j <= m; ++j) {
      if (a[i][j] == '*') hor[i]++;
    }
  }
  for (i = 1; i <= m; ++i) {
    for (j = 1; j <= n; ++j) {
      if (a[j][i] == '*') ver[i]++;
    }
  }
  long long ans = 0;
  for (i = 1; i <= n; ++i) {
    for (j = 1; j <= m; ++j) {
      if (a[i][j] == '*') {
        x = hor[i];
        y = ver[j];
        ans += ((x - 1) * (y - 1));
      }
    }
  }
  cout << ans;
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
long long hor[1001], ver[1001];
signed main() {
  long long n, m;
  cin >> n >> m;
  char a[n + 1][m + 1], z;
  long long i, j, x, y;
  for (i = 1; i <= n; ++i) {
    for (j = 1; j <= m; ++j) cin >> a[i][j];
  }
  for (i = 1; i <= n; ++i) {
    for (j = 1; j <= m; ++j) {
      if (a[i][j] == '*') hor[i]++;
    }
  }
  for (i = 1; i <= m; ++i) {
    for (j = 1; j <= n; ++j) {
      if (a[j][i] == '*') ver[i]++;
    }
  }
  long long ans = 0;
  for (i = 1; i <= n; ++i) {
    for (j = 1; j <= m; ++j) {
      if (a[i][j] == '*') {
        x = hor[i];
        y = ver[j];
        ans += ((x - 1) * (y - 1));
      }
    }
  }
  cout << ans;
  return 0;
}

```