#include <bits/stdc++.h>
using namespace std;
int N, T, M, K;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> N >> M;
  int row[1005], col[1005];
  memset(row, 0, sizeof row);
  memset(col, 0, sizeof col);
  char c;
  char grid[1005][1005];
  for (int(i) = 0; (i) < (N); (i)++) {
    for (int(j) = 0; (j) < (M); (j)++) {
      cin >> c;
      grid[i][j] = c;
      if (c == '*') row[i]++, col[j]++;
    }
  }
  long long ans = 0;
  for (int(i) = 0; (i) < (N); (i)++) {
    for (int(j) = 0; (j) < (M); (j)++) {
      if (grid[i][j] == '*') ans += (row[i] - 1) * (col[j] - 1);
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please formulate 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>
using namespace std;
int N, T, M, K;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> N >> M;
  int row[1005], col[1005];
  memset(row, 0, sizeof row);
  memset(col, 0, sizeof col);
  char c;
  char grid[1005][1005];
  for (int(i) = 0; (i) < (N); (i)++) {
    for (int(j) = 0; (j) < (M); (j)++) {
      cin >> c;
      grid[i][j] = c;
      if (c == '*') row[i]++, col[j]++;
    }
  }
  long long ans = 0;
  for (int(i) = 0; (i) < (N); (i)++) {
    for (int(j) = 0; (j) < (M); (j)++) {
      if (grid[i][j] == '*') ans += (row[i] - 1) * (col[j] - 1);
    }
  }
  cout << ans << endl;
  return 0;
}

```