#include <bits/stdc++.h>
using namespace std;
long long n, m, ans, a[1005];
char c[1005][1005];
vector<long long> v[1005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cin >> c[i][j];
      if (c[i][j] == '*') {
        v[j].push_back(i);
        a[i]++;
      }
    }
  }
  for (int i = 1; i <= m; i++) {
    for (int j = 0; j < v[i].size(); j++) {
      ans += (v[i].size() - 1) * (a[v[i][j]] - 1);
    }
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
long long n, m, ans, a[1005];
char c[1005][1005];
vector<long long> v[1005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cin >> c[i][j];
      if (c[i][j] == '*') {
        v[j].push_back(i);
        a[i]++;
      }
    }
  }
  for (int i = 1; i <= m; i++) {
    for (int j = 0; j < v[i].size(); j++) {
      ans += (v[i].size() - 1) * (a[v[i][j]] - 1);
    }
  }
  cout << ans << '\n';
  return 0;
}

```