#include <bits/stdc++.h>
using namespace ::std;
const int SIZE = 1e3 + 1;
char Mat[SIZE][SIZE];
int TA[SIZE][SIZE];
int sum(int i1, int j1, int i2, int j2) {
  return TA[i2][j2] - TA[i2][j1 - 1] - TA[i1 - 1][j2] + TA[i1 - 1][j1 - 1];
}
int main() {
  int N, M;
  cin >> N >> M;
  for (int i = 1; i <= N; i++)
    for (int j = 1; j <= M; j++) {
      cin >> Mat[i][j];
      TA[i][j] = Mat[i][j] == '*' ? 1 : 0;
      TA[i][j] += TA[i - 1][j] + TA[i][j - 1] - TA[i - 1][j - 1];
    }
  long long sol = 0;
  for (int i = 1; i <= N; i++)
    for (int j = 1; j <= M; j++) {
      if (Mat[i][j] == '*') {
        sol += 1LL * (sum(1, j, N, j) - 1) * (sum(i, 1, i, M) - 1);
      }
    }
  cout << sol << '\n';
  return 0;
}

### Prompt

Your challenge is to write 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;
const int SIZE = 1e3 + 1;
char Mat[SIZE][SIZE];
int TA[SIZE][SIZE];
int sum(int i1, int j1, int i2, int j2) {
  return TA[i2][j2] - TA[i2][j1 - 1] - TA[i1 - 1][j2] + TA[i1 - 1][j1 - 1];
}
int main() {
  int N, M;
  cin >> N >> M;
  for (int i = 1; i <= N; i++)
    for (int j = 1; j <= M; j++) {
      cin >> Mat[i][j];
      TA[i][j] = Mat[i][j] == '*' ? 1 : 0;
      TA[i][j] += TA[i - 1][j] + TA[i][j - 1] - TA[i - 1][j - 1];
    }
  long long sol = 0;
  for (int i = 1; i <= N; i++)
    for (int j = 1; j <= M; j++) {
      if (Mat[i][j] == '*') {
        sol += 1LL * (sum(1, j, N, j) - 1) * (sum(i, 1, i, M) - 1);
      }
    }
  cout << sol << '\n';
  return 0;
}

```