#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
const double eps = 1e-8;
const int nm = 1005;
int n, k, m, t;
long long res;
char s[nm][nm];
int dong[nm], cot[nm];
int main() {
  int i, j, x, y, z, w;
  scanf("%d%d", &n, &m);
  for (i = 0; i < n; i++) scanf("%s", &s[i]);
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++) {
      if (s[i][j] == '*') {
        dong[i]++;
        cot[j]++;
      }
    }
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++) {
      if (s[i][j] == '.') continue;
      res += (dong[i] - 1) * (cot[j] - 1);
    }
  cout << res;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 MOD = 1000000007;
const double eps = 1e-8;
const int nm = 1005;
int n, k, m, t;
long long res;
char s[nm][nm];
int dong[nm], cot[nm];
int main() {
  int i, j, x, y, z, w;
  scanf("%d%d", &n, &m);
  for (i = 0; i < n; i++) scanf("%s", &s[i]);
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++) {
      if (s[i][j] == '*') {
        dong[i]++;
        cot[j]++;
      }
    }
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++) {
      if (s[i][j] == '.') continue;
      res += (dong[i] - 1) * (cot[j] - 1);
    }
  cout << res;
  return 0;
}

```