#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m;
  cin >> n >> m;
  vector<string> s(n);
  for (int i = 0; i < n; ++i) {
    cin >> s[i];
  }
  vector<int> counts1(n, 0), counts2(n, 0);
  int count = 0;
  for (int i = 0; i < n; ++i) {
    if (s[i][0] == '*') {
      ++count;
      ++counts1[i];
    }
  }
  if (count > 1) {
    for (int i = 0; i < n; ++i) {
      if (s[i][0] == '*') {
        counts2[i] += count - 1;
      }
    }
  }
  long long sum_count = 0;
  for (int j = 1; j < m; ++j) {
    int count = 0;
    for (int i = 0; i < n; ++i) {
      if (s[i][j] == '*') {
        ++count;
        sum_count += counts2[i];
      }
    }
    for (int i = 0; i < n; ++i) {
      if (s[i][j] == '*') {
        if (count > 1) {
          counts2[i] += count - 1;
          sum_count += (count - 1) * counts1[i];
        }
        ++counts1[i];
      }
    }
  }
  cout << sum_count;
  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 main() {
  int n, m;
  cin >> n >> m;
  vector<string> s(n);
  for (int i = 0; i < n; ++i) {
    cin >> s[i];
  }
  vector<int> counts1(n, 0), counts2(n, 0);
  int count = 0;
  for (int i = 0; i < n; ++i) {
    if (s[i][0] == '*') {
      ++count;
      ++counts1[i];
    }
  }
  if (count > 1) {
    for (int i = 0; i < n; ++i) {
      if (s[i][0] == '*') {
        counts2[i] += count - 1;
      }
    }
  }
  long long sum_count = 0;
  for (int j = 1; j < m; ++j) {
    int count = 0;
    for (int i = 0; i < n; ++i) {
      if (s[i][j] == '*') {
        ++count;
        sum_count += counts2[i];
      }
    }
    for (int i = 0; i < n; ++i) {
      if (s[i][j] == '*') {
        if (count > 1) {
          counts2[i] += count - 1;
          sum_count += (count - 1) * counts1[i];
        }
        ++counts1[i];
      }
    }
  }
  cout << sum_count;
  return 0;
}

```