#include <bits/stdc++.h>
using namespace std;
long long heng[1100], lie[1100], num[1100][1100], ans, i, j, n, m;
char c;
int main() {
  scanf("%I64d%I64d\n", &n, &m);
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      scanf("%c", &c);
      if (c == '*') {
        num[i][j] = 1;
        heng[i]++;
        lie[j]++;
      }
    }
    scanf("\n");
  }
  for (i = 1; i <= n; i++)
    for (j = 1; j <= m; j++)
      if (num[i][j] == 1) ans += (heng[i] - 1) * (lie[j] - 1);
  printf("%I64d\n", ans);
}

### Prompt

Your task is to create 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 heng[1100], lie[1100], num[1100][1100], ans, i, j, n, m;
char c;
int main() {
  scanf("%I64d%I64d\n", &n, &m);
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      scanf("%c", &c);
      if (c == '*') {
        num[i][j] = 1;
        heng[i]++;
        lie[j]++;
      }
    }
    scanf("\n");
  }
  for (i = 1; i <= n; i++)
    for (j = 1; j <= m; j++)
      if (num[i][j] == 1) ans += (heng[i] - 1) * (lie[j] - 1);
  printf("%I64d\n", ans);
}

```