#include <bits/stdc++.h>
using namespace std;
int a[1000][1000];
int main() {
  int n, m, col[1000];
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    char s[1001];
    cin >> s;
    for (int j = 0; j < m; ++j) a[i][j] = s[j] == '.' ? 0 : 1;
  }
  long long val = 0;
  for (int i = 0; i < m; ++i) {
    col[i] = 0;
    for (int j = 0; j < n; ++j) col[i] += a[j][i];
  }
  for (int i = 0; i < n; ++i) {
    int row = 0;
    for (int j = 0; j < m; ++j) row += a[i][j];
    for (int j = 0; j < m; ++j)
      if (a[i][j]) val += (row - 1) * (col[j] - 1);
  }
  cout << val << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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 a[1000][1000];
int main() {
  int n, m, col[1000];
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    char s[1001];
    cin >> s;
    for (int j = 0; j < m; ++j) a[i][j] = s[j] == '.' ? 0 : 1;
  }
  long long val = 0;
  for (int i = 0; i < m; ++i) {
    col[i] = 0;
    for (int j = 0; j < n; ++j) col[i] += a[j][i];
  }
  for (int i = 0; i < n; ++i) {
    int row = 0;
    for (int j = 0; j < m; ++j) row += a[i][j];
    for (int j = 0; j < m; ++j)
      if (a[i][j]) val += (row - 1) * (col[j] - 1);
  }
  cout << val << endl;
  return 0;
}

```