#include <bits/stdc++.h>
using namespace std;
const int N = 1010, M = N * N;
pair<int, int> p[M];
int r[N];
int n, m, k;
int c[N];
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      char d;
      cin >> d;
      if (d == '*') p[k++] = {i, j}, r[i]++, c[j]++;
    }
    getchar();
  }
  long long res = 0;
  for (int i = 0; i < k; i++) {
    int a = p[i].first, b = p[i].second;
    res += (c[b] - 1) * (r[a] - 1);
  }
  cout << res;
  return 0;
}

### Prompt

Your task is to create 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 N = 1010, M = N * N;
pair<int, int> p[M];
int r[N];
int n, m, k;
int c[N];
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      char d;
      cin >> d;
      if (d == '*') p[k++] = {i, j}, r[i]++, c[j]++;
    }
    getchar();
  }
  long long res = 0;
  for (int i = 0; i < k; i++) {
    int a = p[i].first, b = p[i].second;
    res += (c[b] - 1) * (r[a] - 1);
  }
  cout << res;
  return 0;
}

```