#include <bits/stdc++.h>
using namespace std;
int main(int argc, char *argv[]) {
  int n = 0, m = 0;
  cin >> n >> m;
  int **field = new int *[n];
  for (int i = 0; i < n; ++i) {
    static string row;
    row.reserve(m);
    cin >> row;
    field[i] = new int[m];
    memset(field[i], 0, m * sizeof(int));
    for (int j = 0; j < m; ++j) {
      if (row[j] == '*') {
        field[i][j] = 1;
      }
    }
  }
  long long int count = 0;
  int *col_nums = new int[m];
  memset(col_nums, 0, m * sizeof(int));
  int *row_nums = new int[n];
  memset(row_nums, 0, n * sizeof(int));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (field[i][j] == 1) {
        col_nums[j]++;
        row_nums[i]++;
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (field[i][j] == 1 && col_nums[j] > 0 && row_nums[i]) {
        count += (col_nums[j] - 1) * (row_nums[i] - 1);
      }
    }
  }
  cout << count << endl;
  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 main(int argc, char *argv[]) {
  int n = 0, m = 0;
  cin >> n >> m;
  int **field = new int *[n];
  for (int i = 0; i < n; ++i) {
    static string row;
    row.reserve(m);
    cin >> row;
    field[i] = new int[m];
    memset(field[i], 0, m * sizeof(int));
    for (int j = 0; j < m; ++j) {
      if (row[j] == '*') {
        field[i][j] = 1;
      }
    }
  }
  long long int count = 0;
  int *col_nums = new int[m];
  memset(col_nums, 0, m * sizeof(int));
  int *row_nums = new int[n];
  memset(row_nums, 0, n * sizeof(int));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (field[i][j] == 1) {
        col_nums[j]++;
        row_nums[i]++;
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (field[i][j] == 1 && col_nums[j] > 0 && row_nums[i]) {
        count += (col_nums[j] - 1) * (row_nums[i] - 1);
      }
    }
  }
  cout << count << endl;
  return 0;
}

```