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

### 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>
char map[1005][1005];
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int r = 0; r < n; r++) scanf("%s", map[r]);
  int row[1005] = {0};
  int col[1005] = {0};
  for (int r = 0; r < n; r++) {
    int count = 0;
    for (int c = 0; c < m; c++)
      if (map[r][c] == '*') count++;
    row[r] = count;
  }
  for (int c = 0; c < m; c++) {
    int count = 0;
    for (int r = 0; r < n; r++)
      if (map[r][c] == '*') count++;
    col[c] = count;
  }
  long long answer = 0;
  for (int r = 0; r < n; r++)
    for (int c = 0; c < m; c++)
      if (map[r][c] == '*') answer += (row[r] - 1) * (col[c] - 1);
  printf("%I64d\n", answer);
  return 0;
}

```