#include <bits/stdc++.h>
using namespace std;
char mp;
bool bmp[1000][1000];
long long stb[1000] = {0}, stl[1000] = {0};
int main() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      cin >> mp;
      if (mp == '*') ++stb[j], ++stl[i], bmp[i][j] = true;
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      ans += (bmp[i][j] && stl[i] > 1 ? (stl[i] - 1) * (stb[j] - 1) : 0);
    }
  }
  cout << ans;
  return 0;
}

### 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;
char mp;
bool bmp[1000][1000];
long long stb[1000] = {0}, stl[1000] = {0};
int main() {
  cin.tie(NULL);
  ios_base::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      cin >> mp;
      if (mp == '*') ++stb[j], ++stl[i], bmp[i][j] = true;
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      ans += (bmp[i][j] && stl[i] > 1 ? (stl[i] - 1) * (stb[j] - 1) : 0);
    }
  }
  cout << ans;
  return 0;
}

```