#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
int a, b, c, d, i, j, n, m, k;
char mas[1001][1001];
int cr[1001], cc[1001];
long long ans;
int main() {
  scanf("%d%d", &n, &m);
  for (int _n((n)-1), i(0); i <= _n; i++) scanf("%s", mas[i]);
  for (int _n((n)-1), i(0); i <= _n; i++) {
    for (int _n((m)-1), j(0); j <= _n; j++) {
      if (mas[i][j] == '*') {
        ++cr[i];
        ++cc[j];
      }
    }
  }
  for (int _n((n)-1), i(0); i <= _n; i++) {
    for (int _n((m)-1), j(0); j <= _n; j++) {
      if (mas[i][j] == '*') {
        ans += (long long)(cr[i] - 1) * (cc[j] - 1);
      }
    }
  }
  cout << ans << endl;
}

### 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>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
int a, b, c, d, i, j, n, m, k;
char mas[1001][1001];
int cr[1001], cc[1001];
long long ans;
int main() {
  scanf("%d%d", &n, &m);
  for (int _n((n)-1), i(0); i <= _n; i++) scanf("%s", mas[i]);
  for (int _n((n)-1), i(0); i <= _n; i++) {
    for (int _n((m)-1), j(0); j <= _n; j++) {
      if (mas[i][j] == '*') {
        ++cr[i];
        ++cc[j];
      }
    }
  }
  for (int _n((n)-1), i(0); i <= _n; i++) {
    for (int _n((m)-1), j(0); j <= _n; j++) {
      if (mas[i][j] == '*') {
        ans += (long long)(cr[i] - 1) * (cc[j] - 1);
      }
    }
  }
  cout << ans << endl;
}

```