#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1028;
char field[MAXN][MAXN];
int cnt[MAXN][MAXN][4];
int main() {
  for (int n, m; ~scanf("%d%d", &n, &m);) {
    for (int i = 0; i < n; i++) {
      scanf("%s", field[i]);
    }
    memset(cnt, 0, sizeof cnt);
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        cnt[i][j][0] = i > 0 ? cnt[i - 1][j][0] + (field[i - 1][j] == '*') : 0;
        cnt[i][j][1] = j > 0 ? cnt[i][j - 1][1] + (field[i][j - 1] == '*') : 0;
      }
    }
    for (int i = n - 1; i >= 0; i--) {
      for (int j = m - 1; j >= 0; j--) {
        cnt[i][j][2] =
            i + 1 < n ? cnt[i + 1][j][2] + (field[i + 1][j] == '*') : 0;
        cnt[i][j][3] =
            j + 1 < m ? cnt[i][j + 1][3] + (field[i][j + 1] == '*') : 0;
      }
    }
    long long ans = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        if (field[i][j] == '*') {
          for (int k = 0; k < 4; k++) {
            ans += cnt[i][j][k] * cnt[i][j][(k + 1) % 4];
          }
        }
      }
    }
    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;
const int MAXN = 1028;
char field[MAXN][MAXN];
int cnt[MAXN][MAXN][4];
int main() {
  for (int n, m; ~scanf("%d%d", &n, &m);) {
    for (int i = 0; i < n; i++) {
      scanf("%s", field[i]);
    }
    memset(cnt, 0, sizeof cnt);
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        cnt[i][j][0] = i > 0 ? cnt[i - 1][j][0] + (field[i - 1][j] == '*') : 0;
        cnt[i][j][1] = j > 0 ? cnt[i][j - 1][1] + (field[i][j - 1] == '*') : 0;
      }
    }
    for (int i = n - 1; i >= 0; i--) {
      for (int j = m - 1; j >= 0; j--) {
        cnt[i][j][2] =
            i + 1 < n ? cnt[i + 1][j][2] + (field[i + 1][j] == '*') : 0;
        cnt[i][j][3] =
            j + 1 < m ? cnt[i][j + 1][3] + (field[i][j + 1] == '*') : 0;
      }
    }
    long long ans = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        if (field[i][j] == '*') {
          for (int k = 0; k < 4; k++) {
            ans += cnt[i][j][k] * cnt[i][j][(k + 1) % 4];
          }
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}

```