#include <bits/stdc++.h>
using namespace std;
int n, m;
struct Matrix {
  char c;
  long long up, down, left, right;
} matrix[2000][2000];
long long ans;
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      scanf(" %c", &matrix[i][j].c);
    }
  }
  long long sum;
  for (int j = 1; j <= m; j++) {
    sum = 0;
    for (int i = 1; i <= n; i++) {
      matrix[i][j].up = sum;
      if (matrix[i][j].c == '*') sum += 1;
    }
  }
  for (int j = 1; j <= m; j++) {
    sum = 0;
    for (int i = n; i >= 1; i--) {
      matrix[i][j].down = sum;
      if (matrix[i][j].c == '*') sum += 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    sum = 0;
    for (int j = 1; j <= m; j++) {
      matrix[i][j].left = sum;
      if (matrix[i][j].c == '*') sum += 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    sum = 0;
    for (int j = m; j >= 1; j--) {
      matrix[i][j].right = sum;
      if (matrix[i][j].c == '*') sum += 1;
    }
  }
  ans = 0;
  long long u, d, l, r;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (matrix[i][j].c != '*') continue;
      u = matrix[i][j].up;
      d = matrix[i][j].down;
      l = matrix[i][j].left;
      r = matrix[i][j].right;
      ans += u * l + l * d + d * r + r * u;
    }
  }
  cout << ans << endl;
}

### 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;
int n, m;
struct Matrix {
  char c;
  long long up, down, left, right;
} matrix[2000][2000];
long long ans;
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      scanf(" %c", &matrix[i][j].c);
    }
  }
  long long sum;
  for (int j = 1; j <= m; j++) {
    sum = 0;
    for (int i = 1; i <= n; i++) {
      matrix[i][j].up = sum;
      if (matrix[i][j].c == '*') sum += 1;
    }
  }
  for (int j = 1; j <= m; j++) {
    sum = 0;
    for (int i = n; i >= 1; i--) {
      matrix[i][j].down = sum;
      if (matrix[i][j].c == '*') sum += 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    sum = 0;
    for (int j = 1; j <= m; j++) {
      matrix[i][j].left = sum;
      if (matrix[i][j].c == '*') sum += 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    sum = 0;
    for (int j = m; j >= 1; j--) {
      matrix[i][j].right = sum;
      if (matrix[i][j].c == '*') sum += 1;
    }
  }
  ans = 0;
  long long u, d, l, r;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (matrix[i][j].c != '*') continue;
      u = matrix[i][j].up;
      d = matrix[i][j].down;
      l = matrix[i][j].left;
      r = matrix[i][j].right;
      ans += u * l + l * d + d * r + r * u;
    }
  }
  cout << ans << endl;
}

```