#include <bits/stdc++.h>
using namespace std;
int n, m;
char str[1010][1010];
long long d1[1010][1010], d2[1010][1010];
long long ans;
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) cin >> str[i + 1][j + 1];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) d1[i][j] = d1[i][j - 1] + (str[i][j] == '*');
  for (int j = 1; j <= m; j++)
    for (int i = 1; i <= n; i++) d2[i][j] = d2[i - 1][j] + (str[i][j] == '*');
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (str[i][j] == '.') continue;
      ans += d1[i][j - 1] * d2[i - 1][j];
      ans += (d1[i][m] - d1[i][j]) * d2[i - 1][j];
      ans += (d1[i][m] - d1[i][j]) * (d2[n][j] - d2[i][j]);
      ans += (d2[n][j] - d2[i][j]) * d1[i][j - 1];
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 n, m;
char str[1010][1010];
long long d1[1010][1010], d2[1010][1010];
long long ans;
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) cin >> str[i + 1][j + 1];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) d1[i][j] = d1[i][j - 1] + (str[i][j] == '*');
  for (int j = 1; j <= m; j++)
    for (int i = 1; i <= n; i++) d2[i][j] = d2[i - 1][j] + (str[i][j] == '*');
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (str[i][j] == '.') continue;
      ans += d1[i][j - 1] * d2[i - 1][j];
      ans += (d1[i][m] - d1[i][j]) * d2[i - 1][j];
      ans += (d1[i][m] - d1[i][j]) * (d2[n][j] - d2[i][j]);
      ans += (d2[n][j] - d2[i][j]) * d1[i][j - 1];
    }
  }
  cout << ans;
  return 0;
}

```