#include <bits/stdc++.h>
using namespace std;
const int A = 1e3;
long long int n, m, a[A], b[A], x = -1, sum, y = -1;
string s[A];
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    for (int o = 0; o < m; o++) {
      if (s[i][o] == '*') {
        a[i]++;
        b[o]++;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (s[i][j] == '*') {
        sum += (a[i] - 1) * (b[j] - 1);
      }
    }
  }
  cout << sum;
}

### Prompt

Your challenge is to write 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;
const int A = 1e3;
long long int n, m, a[A], b[A], x = -1, sum, y = -1;
string s[A];
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    for (int o = 0; o < m; o++) {
      if (s[i][o] == '*') {
        a[i]++;
        b[o]++;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (s[i][j] == '*') {
        sum += (a[i] - 1) * (b[j] - 1);
      }
    }
  }
  cout << sum;
}

```