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

### Prompt

Your task is to create 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 m, n;
  cin >> n >> m;
  cin.get();
  vector<string> map;
  for (int i = 0; i < n; i++) {
    string s;
    getline(cin, s);
    map.push_back(s);
  }
  vector<int> rowCount(n, 0);
  vector<int> colCount(m, 0);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (map[i][j] == '*') {
        rowCount[i]++;
        colCount[j]++;
      }
    }
  }
  unsigned long long count = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      if (map[i][j] == '*') count += (rowCount[i] - 1) * (colCount[j] - 1);
    }
  }
  cout << count;
  return 0;
}

```