#include <bits/stdc++.h>
using namespace std;
const int N = 1003;
int m, n, row[2][N][N], col[2][N][N];
char a[N][N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> m >> n;
  for (int i = 1; i <= m; ++i)
    for (int j = 1; j <= n; ++j) {
      cin >> a[i][j];
      row[0][i][j] = row[0][i][j - 1] + (a[i][j] == '*');
      col[0][i][j] = col[0][i - 1][j] + (a[i][j] == '*');
    }
  for (int i = m; i > 0; --i)
    for (int j = n; j > 0; --j) {
      row[1][i][j] = row[1][i][j + 1] + (a[i][j] == '*');
      col[1][i][j] = col[1][i + 1][j] + (a[i][j] == '*');
    }
  long long ans = 0;
  for (int i = 1; i <= m; ++i)
    for (int j = 1; j <= n; ++j)
      if (a[i][j] == '*')
        ans += (row[0][i][j] - 1) * (col[0][i][j] - 1) +
               (row[0][i][j] - 1) * (col[1][i][j] - 1) +
               (row[1][i][j] - 1) * (col[0][i][j] - 1) +
               (row[1][i][j] - 1) * (col[1][i][j] - 1);
  cout << ans;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 int N = 1003;
int m, n, row[2][N][N], col[2][N][N];
char a[N][N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> m >> n;
  for (int i = 1; i <= m; ++i)
    for (int j = 1; j <= n; ++j) {
      cin >> a[i][j];
      row[0][i][j] = row[0][i][j - 1] + (a[i][j] == '*');
      col[0][i][j] = col[0][i - 1][j] + (a[i][j] == '*');
    }
  for (int i = m; i > 0; --i)
    for (int j = n; j > 0; --j) {
      row[1][i][j] = row[1][i][j + 1] + (a[i][j] == '*');
      col[1][i][j] = col[1][i + 1][j] + (a[i][j] == '*');
    }
  long long ans = 0;
  for (int i = 1; i <= m; ++i)
    for (int j = 1; j <= n; ++j)
      if (a[i][j] == '*')
        ans += (row[0][i][j] - 1) * (col[0][i][j] - 1) +
               (row[0][i][j] - 1) * (col[1][i][j] - 1) +
               (row[1][i][j] - 1) * (col[0][i][j] - 1) +
               (row[1][i][j] - 1) * (col[1][i][j] - 1);
  cout << ans;
}

```