#include <bits/stdc++.h>
using namespace std;
int up[1001][1001];
int down[1001][1001];
int lft[1001][1001];
int rgt[1001][1001];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  memset(up, 0, sizeof up);
  memset(down, 0, sizeof down);
  memset(lft, 0, sizeof lft);
  memset(rgt, 0, sizeof rgt);
  vector<string> vs(n);
  for (int i = 0; i < n; ++i) {
    cin >> vs[i];
  }
  for (int i = 0; i < n; ++i) {
    int st = 0;
    for (int j = 0; j < m; ++j) {
      lft[i][j] = st;
      if (vs[i][j] == '*') {
        ++st;
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    int st = 0;
    for (int j = m - 1; j >= 0; --j) {
      rgt[i][j] = st;
      if (vs[i][j] == '*') {
        ++st;
      }
    }
  }
  for (int j = 0; j < m; ++j) {
    int st = 0;
    for (int i = 0; i < n; ++i) {
      up[i][j] = st;
      if (vs[i][j] == '*') {
        ++st;
      }
    }
  }
  for (int j = 0; j < m; ++j) {
    int st = 0;
    for (int i = n - 1; i >= 0; --i) {
      down[i][j] = st;
      if (vs[i][j] == '*') {
        ++st;
      }
    }
  }
  long long int ans = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (vs[i][j] == '*') {
        ans += lft[i][j] * up[i][j];
        ans += lft[i][j] * down[i][j];
        ans += rgt[i][j] * up[i][j];
        ans += rgt[i][j] * down[i][j];
      }
    }
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Develop a solution in CPP 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;
int up[1001][1001];
int down[1001][1001];
int lft[1001][1001];
int rgt[1001][1001];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  memset(up, 0, sizeof up);
  memset(down, 0, sizeof down);
  memset(lft, 0, sizeof lft);
  memset(rgt, 0, sizeof rgt);
  vector<string> vs(n);
  for (int i = 0; i < n; ++i) {
    cin >> vs[i];
  }
  for (int i = 0; i < n; ++i) {
    int st = 0;
    for (int j = 0; j < m; ++j) {
      lft[i][j] = st;
      if (vs[i][j] == '*') {
        ++st;
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    int st = 0;
    for (int j = m - 1; j >= 0; --j) {
      rgt[i][j] = st;
      if (vs[i][j] == '*') {
        ++st;
      }
    }
  }
  for (int j = 0; j < m; ++j) {
    int st = 0;
    for (int i = 0; i < n; ++i) {
      up[i][j] = st;
      if (vs[i][j] == '*') {
        ++st;
      }
    }
  }
  for (int j = 0; j < m; ++j) {
    int st = 0;
    for (int i = n - 1; i >= 0; --i) {
      down[i][j] = st;
      if (vs[i][j] == '*') {
        ++st;
      }
    }
  }
  long long int ans = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (vs[i][j] == '*') {
        ans += lft[i][j] * up[i][j];
        ans += lft[i][j] * down[i][j];
        ans += rgt[i][j] * up[i][j];
        ans += rgt[i][j] * down[i][j];
      }
    }
  }
  cout << ans << '\n';
  return 0;
}

```