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

### 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;
char s[1005][1005];
int numx[1005], numy[1005], n, m;
int main() {
  long long ans = 0;
  int i, j;
  scanf("%d %d", &n, &m);
  for (i = 0; i < n; i++) scanf("%s", s[i]);
  memset(numx, 0, sizeof(numx));
  memset(numy, 0, sizeof(numy));
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++)
      if (s[i][j] == '*') {
        numx[i]++;
        numy[j]++;
      }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++)
      if (s[i][j] == '*') {
        long long temp = (long long)(numx[i] - 1) * (long long)(numy[j] - 1);
        ans += temp;
      }
  }
  printf("%I64d", ans);
  return 0;
}

```