#include <bits/stdc++.h>
using namespace std;
long long col[1024][1024], row[1024][1024];
int m, n;
char buffer[1024];
char G[1024][1024];
int main() {
  scanf("%d %d", &m, &n);
  for (int i = 0; i < (m); i++) {
    scanf("%s", buffer);
    for (int j = 0; j < (n); j++) {
      row[i][j] = j ? row[i][j - 1] : 0;
      col[i][j] = i ? col[i - 1][j] : 0;
      row[i][j] += buffer[j] == '*';
      col[i][j] += buffer[j] == '*';
    }
    strcpy(G[i], buffer);
  }
  long long ans = 0;
  for (int i = 0; i < (m); i++)
    for (int j = 0; j < (n); j++)
      if (G[i][j] == '*') ans += (row[i][n - 1] - 1) * (col[m - 1][j] - 1);
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 col[1024][1024], row[1024][1024];
int m, n;
char buffer[1024];
char G[1024][1024];
int main() {
  scanf("%d %d", &m, &n);
  for (int i = 0; i < (m); i++) {
    scanf("%s", buffer);
    for (int j = 0; j < (n); j++) {
      row[i][j] = j ? row[i][j - 1] : 0;
      col[i][j] = i ? col[i - 1][j] : 0;
      row[i][j] += buffer[j] == '*';
      col[i][j] += buffer[j] == '*';
    }
    strcpy(G[i], buffer);
  }
  long long ans = 0;
  for (int i = 0; i < (m); i++)
    for (int j = 0; j < (n); j++)
      if (G[i][j] == '*') ans += (row[i][n - 1] - 1) * (col[m - 1][j] - 1);
  cout << ans << endl;
  return 0;
}

```