#include <bits/stdc++.h>
using namespace std;
inline void print(long long x) {
  printf("%lld", x);
  puts("");
}
int mp[1005][1005], row[1005], col[1005];
int dp[1005][1005];
int main() {
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    memset(mp, 0, sizeof(mp));
    memset(row, 0, sizeof(row));
    memset(col, 0, sizeof(col));
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) {
        char ch;
        cin >> ch;
        mp[i][j] = (ch == '*');
        row[i] += mp[i][j];
        col[j] += mp[i][j];
      }
    long long ans = 0;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++)
        if (mp[i][j]) ans += (row[i] - 1) * (col[j] - 1);
    print(ans);
  }
  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;
inline void print(long long x) {
  printf("%lld", x);
  puts("");
}
int mp[1005][1005], row[1005], col[1005];
int dp[1005][1005];
int main() {
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    memset(mp, 0, sizeof(mp));
    memset(row, 0, sizeof(row));
    memset(col, 0, sizeof(col));
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) {
        char ch;
        cin >> ch;
        mp[i][j] = (ch == '*');
        row[i] += mp[i][j];
        col[j] += mp[i][j];
      }
    long long ans = 0;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++)
        if (mp[i][j]) ans += (row[i] - 1) * (col[j] - 1);
    print(ans);
  }
  return 0;
}

```