#include <bits/stdc++.h>
using namespace std;
int n, m;
long long l[1005][1005], r[1005][1005], u[1005][1005], d[1005][1005];
char s[1005][1005];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%s", s[i] + 1);
  for (int i = 1; i <= n; i++) {
    for (int j = m; j >= 1; j--) {
      r[i][j] = r[i][j + 1] + (s[i][j] == '*');
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      l[i][j] = l[i][j - 1] + (s[i][j] == '*');
    }
  }
  for (int j = 1; j <= m; j++) {
    for (int i = 1; i <= n; i++) {
      u[i][j] = u[i - 1][j] + (s[i][j] == '*');
    }
  }
  for (int j = 1; j <= m; j++) {
    for (int i = n; i >= 1; i--) {
      d[i][j] = d[i + 1][j] + (s[i][j] == '*');
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (s[i][j] == '*') {
        ans += (l[i][j] - 1) * (u[i][j] - 1);
        ans += (r[i][j] - 1) * (u[i][j] - 1);
        ans += (l[i][j] - 1) * (d[i][j] - 1);
        ans += (r[i][j] - 1) * (d[i][j] - 1);
      }
    }
  }
  printf("%I64d\n", ans);
}

### Prompt

Please create a solution in cpp 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 n, m;
long long l[1005][1005], r[1005][1005], u[1005][1005], d[1005][1005];
char s[1005][1005];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%s", s[i] + 1);
  for (int i = 1; i <= n; i++) {
    for (int j = m; j >= 1; j--) {
      r[i][j] = r[i][j + 1] + (s[i][j] == '*');
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      l[i][j] = l[i][j - 1] + (s[i][j] == '*');
    }
  }
  for (int j = 1; j <= m; j++) {
    for (int i = 1; i <= n; i++) {
      u[i][j] = u[i - 1][j] + (s[i][j] == '*');
    }
  }
  for (int j = 1; j <= m; j++) {
    for (int i = n; i >= 1; i--) {
      d[i][j] = d[i + 1][j] + (s[i][j] == '*');
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (s[i][j] == '*') {
        ans += (l[i][j] - 1) * (u[i][j] - 1);
        ans += (r[i][j] - 1) * (u[i][j] - 1);
        ans += (l[i][j] - 1) * (d[i][j] - 1);
        ans += (r[i][j] - 1) * (d[i][j] - 1);
      }
    }
  }
  printf("%I64d\n", ans);
}

```