#include <bits/stdc++.h>
using namespace std;
long long dp[1010][1010];
int main() {
  int n, m;
  cin >> n >> m;
  string v[n];
  for (int i = 0; i < n; i++) cin >> v[i];
  for (int i = 0; i < 1010; i++)
    for (int j = 0; j < 1010; j++) dp[i][j] = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (v[i][j] == '*') dp[i + 1][j + 1] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
      dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1] + dp[i][j];
  long long res = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (v[i][j] == '*') {
        long long c1 = (dp[i + 1][m - 1 + 1] - dp[i + 1][j] - dp[i][m - 1 + 1] +
                        dp[i][j]) -
                       1;
        long long c2 = (dp[n - 1 + 1][j + 1] - dp[n - 1 + 1][j] - dp[i][j + 1] +
                        dp[i][j]) -
                       1;
        long long c3 =
            (dp[i + 1][j + 1] - dp[i + 1][0] - dp[i][j + 1] + dp[i][0]) - 1;
        long long c4 =
            (dp[i + 1][j + 1] - dp[i + 1][j] - dp[0][j + 1] + dp[0][j]) - 1;
        res += (c1 * c2);
        res += (c1 * c4);
        res += (c3 * c2);
        res += (c3 * c4);
      }
    }
  }
  cout << res << 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;
long long dp[1010][1010];
int main() {
  int n, m;
  cin >> n >> m;
  string v[n];
  for (int i = 0; i < n; i++) cin >> v[i];
  for (int i = 0; i < 1010; i++)
    for (int j = 0; j < 1010; j++) dp[i][j] = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (v[i][j] == '*') dp[i + 1][j + 1] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
      dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1] + dp[i][j];
  long long res = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (v[i][j] == '*') {
        long long c1 = (dp[i + 1][m - 1 + 1] - dp[i + 1][j] - dp[i][m - 1 + 1] +
                        dp[i][j]) -
                       1;
        long long c2 = (dp[n - 1 + 1][j + 1] - dp[n - 1 + 1][j] - dp[i][j + 1] +
                        dp[i][j]) -
                       1;
        long long c3 =
            (dp[i + 1][j + 1] - dp[i + 1][0] - dp[i][j + 1] + dp[i][0]) - 1;
        long long c4 =
            (dp[i + 1][j + 1] - dp[i + 1][j] - dp[0][j + 1] + dp[0][j]) - 1;
        res += (c1 * c2);
        res += (c1 * c4);
        res += (c3 * c2);
        res += (c3 * c4);
      }
    }
  }
  cout << res << endl;
  return 0;
}

```