#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  string mat[n];
  vector<vector<int> > rows(n, vector<int>(m, 0));
  vector<vector<int> > cols(n, vector<int>(m, 0));
  for (int i = 0; i < n; ++i) {
    cin >> mat[i];
    for (int j = 0; j < m; ++j) {
      if (mat[i][j] == '*') {
        ++rows[i][j];
        ++cols[i][j];
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 1; j < m; ++j) {
      cols[i][j] += cols[i][j - 1];
    }
  }
  for (int j = 0; j < m; ++j) {
    for (int i = 1; i < n; ++i) {
      rows[i][j] += rows[i - 1][j];
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (mat[i][j] == '*') {
        long long up = i - 1 >= 0 ? rows[i - 1][j] : 0;
        long long down = rows[n - 1][j] - rows[i][j];
        long long left = j - 1 >= 0 ? cols[i][j - 1] : 0;
        ;
        long long right = cols[i][m - 1] - cols[i][j];
        ans += (left + right) * (up + down);
      }
    }
  }
  cout << ans << "\n";
}

### 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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  string mat[n];
  vector<vector<int> > rows(n, vector<int>(m, 0));
  vector<vector<int> > cols(n, vector<int>(m, 0));
  for (int i = 0; i < n; ++i) {
    cin >> mat[i];
    for (int j = 0; j < m; ++j) {
      if (mat[i][j] == '*') {
        ++rows[i][j];
        ++cols[i][j];
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 1; j < m; ++j) {
      cols[i][j] += cols[i][j - 1];
    }
  }
  for (int j = 0; j < m; ++j) {
    for (int i = 1; i < n; ++i) {
      rows[i][j] += rows[i - 1][j];
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (mat[i][j] == '*') {
        long long up = i - 1 >= 0 ? rows[i - 1][j] : 0;
        long long down = rows[n - 1][j] - rows[i][j];
        long long left = j - 1 >= 0 ? cols[i][j - 1] : 0;
        ;
        long long right = cols[i][m - 1] - cols[i][j];
        ans += (left + right) * (up + down);
      }
    }
  }
  cout << ans << "\n";
}

```