#include <bits/stdc++.h>
using namespace std;
int row[1005], col[1005], ar[1005][1005];
int N, M;
int main() {
  cin >> N >> M;
  string str;
  for (int i = 0; i < N; i++) {
    cin >> str;
    for (int j = 0; j < M; j++)
      if (str[j] == '*') {
        row[i]++;
        col[j]++;
        ar[i][j] = 1;
      }
  }
  long long ans = 0ll;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < M; j++) {
      if (!ar[i][j]) {
        continue;
      }
      ans += ((row[i] - 1) * (col[j] - 1));
    }
  }
  cout << ans << endl;
  return (0);
}

### Prompt

Please create a solution in CPP 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 row[1005], col[1005], ar[1005][1005];
int N, M;
int main() {
  cin >> N >> M;
  string str;
  for (int i = 0; i < N; i++) {
    cin >> str;
    for (int j = 0; j < M; j++)
      if (str[j] == '*') {
        row[i]++;
        col[j]++;
        ar[i][j] = 1;
      }
  }
  long long ans = 0ll;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < M; j++) {
      if (!ar[i][j]) {
        continue;
      }
      ans += ((row[i] - 1) * (col[j] - 1));
    }
  }
  cout << ans << endl;
  return (0);
}

```