#include <bits/stdc++.h>
using namespace std;
char c[1111][1111];
long long row[1111], col[1111];
int main(void) {
  int i, j, n, m;
  long long res = 0;
  scanf("%d %d", &n, &m);
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++) {
      scanf(" %c", &c[j][i]);
      col[j] += (c[j][i] == '*');
      row[i] += (c[j][i] == '*');
    }
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++)
      if (c[j][i] == '*') res += (col[j] - 1) * (row[i] - 1);
  printf("%I64d\n", res);
  return 0;
}

### Prompt

In cpp, your task is to solve 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 c[1111][1111];
long long row[1111], col[1111];
int main(void) {
  int i, j, n, m;
  long long res = 0;
  scanf("%d %d", &n, &m);
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++) {
      scanf(" %c", &c[j][i]);
      col[j] += (c[j][i] == '*');
      row[i] += (c[j][i] == '*');
    }
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++)
      if (c[j][i] == '*') res += (col[j] - 1) * (row[i] - 1);
  printf("%I64d\n", res);
  return 0;
}

```