#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e3 + 10;
long long n, m, b[MAXN][MAXN][5], ans;
char a[MAXN][MAXN];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < MAXN; i++)
    for (int j = 0; j < MAXN; j++) a[i][j] = '.';
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) cin >> a[i][j];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
      b[i][j][1] = b[i - 1][j][1] + (a[i - 1][j] == '*'),
      b[i][j][4] = b[i][j - 1][4] + (a[i][j - 1] == '*');
  for (int i = n; i > 0; i--)
    for (int j = m; j > 0; j--)
      b[i][j][2] = b[i][j + 1][2] + (a[i][j + 1] == '*'),
      b[i][j][3] = b[i + 1][j][3] + (a[i + 1][j] == '*');
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      if (a[i][j] == '.') continue;
      long long t = b[i][j][1] * b[i][j][2] + b[i][j][2] * b[i][j][3] +
                    b[i][j][3] * b[i][j][4] + b[i][j][4] * b[i][j][1];
      ans += t;
    }
  cout << ans;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 MAXN = 1e3 + 10;
long long n, m, b[MAXN][MAXN][5], ans;
char a[MAXN][MAXN];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < MAXN; i++)
    for (int j = 0; j < MAXN; j++) a[i][j] = '.';
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) cin >> a[i][j];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
      b[i][j][1] = b[i - 1][j][1] + (a[i - 1][j] == '*'),
      b[i][j][4] = b[i][j - 1][4] + (a[i][j - 1] == '*');
  for (int i = n; i > 0; i--)
    for (int j = m; j > 0; j--)
      b[i][j][2] = b[i][j + 1][2] + (a[i][j + 1] == '*'),
      b[i][j][3] = b[i + 1][j][3] + (a[i + 1][j] == '*');
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) {
      if (a[i][j] == '.') continue;
      long long t = b[i][j][1] * b[i][j][2] + b[i][j][2] * b[i][j][3] +
                    b[i][j][3] * b[i][j][4] + b[i][j][4] * b[i][j][1];
      ans += t;
    }
  cout << ans;
  return 0;
}

```