#include <bits/stdc++.h>
using namespace std;
char c[1201][1201];
long long pref[1201][1201], n, m, i, j, k1, k2, k3, k4, ans;
long long get(long long abcd, long long ABCD, long long x2, long long y2) {
  return pref[x2][y2] - pref[abcd - 1][y2] - pref[x2][ABCD - 1] +
         pref[abcd - 1][ABCD - 1];
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= m; j++) {
      cin >> c[i][j];
      pref[i][j] = pref[i - 1][j] + pref[i][j - 1] - pref[i - 1][j - 1] +
                   (c[i][j] == '*');
    }
  for (i = 1; i <= n; i++)
    for (j = 1; j <= m; j++)
      if (c[i][j] == '*') {
        k1 = get(i, 1, i, j - 1);
        k2 = get(1, j, i - 1, j);
        k3 = get(i, j + 1, i, m);
        k4 = get(i + 1, j, n, j);
        ans += k1 * k2 + k2 * k3 + k3 * k4 + k4 * k1;
      }
  cout << ans << '\n';
  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;
char c[1201][1201];
long long pref[1201][1201], n, m, i, j, k1, k2, k3, k4, ans;
long long get(long long abcd, long long ABCD, long long x2, long long y2) {
  return pref[x2][y2] - pref[abcd - 1][y2] - pref[x2][ABCD - 1] +
         pref[abcd - 1][ABCD - 1];
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= m; j++) {
      cin >> c[i][j];
      pref[i][j] = pref[i - 1][j] + pref[i][j - 1] - pref[i - 1][j - 1] +
                   (c[i][j] == '*');
    }
  for (i = 1; i <= n; i++)
    for (j = 1; j <= m; j++)
      if (c[i][j] == '*') {
        k1 = get(i, 1, i, j - 1);
        k2 = get(1, j, i - 1, j);
        k3 = get(i, j + 1, i, m);
        k4 = get(i + 1, j, n, j);
        ans += k1 * k2 + k2 * k3 + k3 * k4 + k4 * k1;
      }
  cout << ans << '\n';
  return 0;
}

```