#include <bits/stdc++.h>
using namespace std;
const long long N = 1010;
string s[N];
long long A[N][N], B[N][N];
int32_t main() {
  long long n, m, ans = 0;
  cin >> n >> m;
  for (long long i = 0; i < n; i++) cin >> s[i];
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= m; j++) {
      A[i][j] += A[i][j - 1] + (s[i - 1][j - 1] == '*');
      B[i][j] += B[i - 1][j] + (s[i - 1][j - 1] == '*');
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= m; j++) {
      if (s[i - 1][j - 1] == '*') {
        ans += (A[i][m] - 1) * (B[n][j] - 1);
      }
    }
  }
  cout << ans << endl;
}

### 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;
const long long N = 1010;
string s[N];
long long A[N][N], B[N][N];
int32_t main() {
  long long n, m, ans = 0;
  cin >> n >> m;
  for (long long i = 0; i < n; i++) cin >> s[i];
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= m; j++) {
      A[i][j] += A[i][j - 1] + (s[i - 1][j - 1] == '*');
      B[i][j] += B[i - 1][j] + (s[i - 1][j - 1] == '*');
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= m; j++) {
      if (s[i - 1][j - 1] == '*') {
        ans += (A[i][m] - 1) * (B[n][j] - 1);
      }
    }
  }
  cout << ans << endl;
}

```