#include <bits/stdc++.h>
using namespace std;
int rlCNT[1000][1000];
int rrCNT[1000][1000];
int cuCNT[1000][1000];
int cdCNT[1000][1000];
char grid[1000][1000];
int n, m;
int main() {
  while (cin >> n >> m) {
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < m; ++j) cin >> grid[i][j];
    memset(rlCNT, 0, sizeof rlCNT);
    memset(rrCNT, 0, sizeof rrCNT);
    memset(cuCNT, 0, sizeof cuCNT);
    memset(cdCNT, 0, sizeof cdCNT);
    for (int i = 0; i < n; ++i) {
      rlCNT[i][0] = (grid[i][0] == '*');
      for (int j = 1; j < m; ++j)
        rlCNT[i][j] = rlCNT[i][j - 1] + (grid[i][j] == '*');
    }
    for (int i = 0; i < n; ++i) {
      rrCNT[i][m - 1] = grid[i][m - 1] == '*';
      for (int j = m - 2; j >= 0; --j)
        rrCNT[i][j] = rrCNT[i][j + 1] + (grid[i][j] == '*');
    }
    for (int j = 0; j < m; ++j) {
      cuCNT[0][j] = grid[0][j] == '*';
      for (int i = 1; i < n; ++i)
        cuCNT[i][j] = cuCNT[i - 1][j] + (grid[i][j] == '*');
    }
    for (int j = 0; j < m; ++j) {
      cdCNT[n - 1][j] = grid[n - 1][j] == '*';
      for (int i = n - 2; i >= 0; --i)
        cdCNT[i][j] = cdCNT[i + 1][j] + (grid[i][j] == '*');
    }
    unsigned long long res = 0;
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < m; ++j) {
        if (grid[i][j] == '*') {
          res += (cdCNT[i][j] - 1) * (rrCNT[i][j] - 1) +
                 (cdCNT[i][j] - 1) * (rlCNT[i][j] - 1) +
                 (cuCNT[i][j] - 1) * (rlCNT[i][j] - 1) +
                 (cuCNT[i][j] - 1) * (rrCNT[i][j] - 1);
        }
      }
    cout << res << endl;
  }
}

### Prompt

Please formulate 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 rlCNT[1000][1000];
int rrCNT[1000][1000];
int cuCNT[1000][1000];
int cdCNT[1000][1000];
char grid[1000][1000];
int n, m;
int main() {
  while (cin >> n >> m) {
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < m; ++j) cin >> grid[i][j];
    memset(rlCNT, 0, sizeof rlCNT);
    memset(rrCNT, 0, sizeof rrCNT);
    memset(cuCNT, 0, sizeof cuCNT);
    memset(cdCNT, 0, sizeof cdCNT);
    for (int i = 0; i < n; ++i) {
      rlCNT[i][0] = (grid[i][0] == '*');
      for (int j = 1; j < m; ++j)
        rlCNT[i][j] = rlCNT[i][j - 1] + (grid[i][j] == '*');
    }
    for (int i = 0; i < n; ++i) {
      rrCNT[i][m - 1] = grid[i][m - 1] == '*';
      for (int j = m - 2; j >= 0; --j)
        rrCNT[i][j] = rrCNT[i][j + 1] + (grid[i][j] == '*');
    }
    for (int j = 0; j < m; ++j) {
      cuCNT[0][j] = grid[0][j] == '*';
      for (int i = 1; i < n; ++i)
        cuCNT[i][j] = cuCNT[i - 1][j] + (grid[i][j] == '*');
    }
    for (int j = 0; j < m; ++j) {
      cdCNT[n - 1][j] = grid[n - 1][j] == '*';
      for (int i = n - 2; i >= 0; --i)
        cdCNT[i][j] = cdCNT[i + 1][j] + (grid[i][j] == '*');
    }
    unsigned long long res = 0;
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < m; ++j) {
        if (grid[i][j] == '*') {
          res += (cdCNT[i][j] - 1) * (rrCNT[i][j] - 1) +
                 (cdCNT[i][j] - 1) * (rlCNT[i][j] - 1) +
                 (cuCNT[i][j] - 1) * (rlCNT[i][j] - 1) +
                 (cuCNT[i][j] - 1) * (rrCNT[i][j] - 1);
        }
      }
    cout << res << endl;
  }
}

```