#include <bits/stdc++.h>
using namespace std;
char grid[1000][1000];
int row[1000];
int col[1000];
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < 1000; i++) {
    row[i] = 0;
    col[i] = 0;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> grid[i][j];
      if (grid[i][j] == '*') {
        row[i]++;
        col[j]++;
      }
    }
  }
  int inrow, incol;
  long long tri = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (grid[i][j] != '*') {
        continue;
      }
      inrow = row[i];
      incol = col[j];
      if (incol >= 2 && inrow >= 2) {
        tri += (incol - 1) * (inrow - 1);
      }
    }
  }
  cout << tri << endl;
  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>
using namespace std;
char grid[1000][1000];
int row[1000];
int col[1000];
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < 1000; i++) {
    row[i] = 0;
    col[i] = 0;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> grid[i][j];
      if (grid[i][j] == '*') {
        row[i]++;
        col[j]++;
      }
    }
  }
  int inrow, incol;
  long long tri = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (grid[i][j] != '*') {
        continue;
      }
      inrow = row[i];
      incol = col[j];
      if (incol >= 2 && inrow >= 2) {
        tri += (incol - 1) * (inrow - 1);
      }
    }
  }
  cout << tri << endl;
  return 0;
}

```