#include <bits/stdc++.h>
int row[1005], col[1005];
char field[1005][1005], buf[1005];
int main(void) {
  int i, j, m, n;
  long long total;
  scanf("%d %d", &n, &m);
  for (i = 0; i < n; ++i) {
    scanf("%s", buf);
    for (j = 0; j < m; ++j) {
      field[i][j] = buf[j];
      if (field[i][j] == '*') {
        row[i]++;
        col[j]++;
      }
    }
  }
  total = 0;
  for (i = 0; i < n; ++i)
    for (j = 0; j < m; ++j)
      if (field[i][j] == '*') total += (row[i] - 1) * (col[j] - 1);
  printf("%I64d\n", total);
  return 0;
}

### Prompt

Create a solution in CPP for 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>
int row[1005], col[1005];
char field[1005][1005], buf[1005];
int main(void) {
  int i, j, m, n;
  long long total;
  scanf("%d %d", &n, &m);
  for (i = 0; i < n; ++i) {
    scanf("%s", buf);
    for (j = 0; j < m; ++j) {
      field[i][j] = buf[j];
      if (field[i][j] == '*') {
        row[i]++;
        col[j]++;
      }
    }
  }
  total = 0;
  for (i = 0; i < n; ++i)
    for (j = 0; j < m; ++j)
      if (field[i][j] == '*') total += (row[i] - 1) * (col[j] - 1);
  printf("%I64d\n", total);
  return 0;
}

```