#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m;
  cin >> n >> m;
  long long ans = 0;
  vector<string> arr(n);
  for (int i = 0; i < n; i++) cin >> arr[i];
  vector<long long> pf1(n), pf2(m);
  for (int i = 0; i < n; i++) {
    long long sum = 0;
    for (int j = 0; j < m; j++) {
      if (arr[i][j] == '*') sum++;
    }
    pf1[i] = sum;
  }
  for (int i = 0; i < m; i++) {
    long long sum = 0;
    for (int j = 0; j < n; j++) {
      if (arr[j][i] == '*') sum++;
    }
    pf2[i] = sum;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (arr[i][j] == '*') ans = ans + (pf1[i] - 1) * (pf2[j] - 1);
    }
  }
  cout << ans;
  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 main() {
  int n, m;
  cin >> n >> m;
  long long ans = 0;
  vector<string> arr(n);
  for (int i = 0; i < n; i++) cin >> arr[i];
  vector<long long> pf1(n), pf2(m);
  for (int i = 0; i < n; i++) {
    long long sum = 0;
    for (int j = 0; j < m; j++) {
      if (arr[i][j] == '*') sum++;
    }
    pf1[i] = sum;
  }
  for (int i = 0; i < m; i++) {
    long long sum = 0;
    for (int j = 0; j < n; j++) {
      if (arr[j][i] == '*') sum++;
    }
    pf2[i] = sum;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (arr[i][j] == '*') ans = ans + (pf1[i] - 1) * (pf2[j] - 1);
    }
  }
  cout << ans;
  return 0;
}

```