#include <bits/stdc++.h>
using namespace std;
int M[1002][1002];
int L[1002][1002];
int U[1002][1002];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) {
      char c;
      cin >> c;
      M[i][j] = c == '*';
      L[i][j] = L[i][j - 1];
      U[i][j] = U[i - 1][j];
      if (M[i][j]) ++L[i][j], ++U[i][j];
    }
  long long ans = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j)
      if (M[i][j]) {
        ans += (L[i][j] - 1) * (U[i][j] - 1);
        ans += (L[i][m] - L[i][j]) * (U[i][j] - 1);
        ans += (L[i][j] - 1) * (U[n][j] - U[i][j]);
        ans += (L[i][m] - L[i][j]) * (U[n][j] - U[i][j]);
      }
  cout << ans << endl;
  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;
int M[1002][1002];
int L[1002][1002];
int U[1002][1002];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) {
      char c;
      cin >> c;
      M[i][j] = c == '*';
      L[i][j] = L[i][j - 1];
      U[i][j] = U[i - 1][j];
      if (M[i][j]) ++L[i][j], ++U[i][j];
    }
  long long ans = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j)
      if (M[i][j]) {
        ans += (L[i][j] - 1) * (U[i][j] - 1);
        ans += (L[i][m] - L[i][j]) * (U[i][j] - 1);
        ans += (L[i][j] - 1) * (U[n][j] - U[i][j]);
        ans += (L[i][m] - L[i][j]) * (U[n][j] - U[i][j]);
      }
  cout << ans << endl;
  return 0;
}

```