#include <bits/stdc++.h>
using namespace std;
int main() {
  long long result = 0;
  int x, y;
  cin >> x >> y;
  int col[1000];
  int row[1000];
  for (int i = 0; i < y; i++) {
    col[i] = 0;
  }
  for (int i = 0; i < x; i++) {
    row[i] = 0;
  }
  wstring table[1000];
  for (int i = 0; i < x; i++) {
    wstring line;
    wcin >> line;
    table[i] = line;
    for (int e = 0; e < y; e++) {
      if (line[e] == '*') {
        col[e] += 1;
        row[i] += 1;
      }
    }
  }
  for (int i = 0; i < x; i++) {
    for (int e = 0; e < y; e++) {
      if (table[i][e] == '*') {
        result += (row[i] - 1) * (col[e] - 1);
      }
    }
  }
  cout << result;
  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 main() {
  long long result = 0;
  int x, y;
  cin >> x >> y;
  int col[1000];
  int row[1000];
  for (int i = 0; i < y; i++) {
    col[i] = 0;
  }
  for (int i = 0; i < x; i++) {
    row[i] = 0;
  }
  wstring table[1000];
  for (int i = 0; i < x; i++) {
    wstring line;
    wcin >> line;
    table[i] = line;
    for (int e = 0; e < y; e++) {
      if (line[e] == '*') {
        col[e] += 1;
        row[i] += 1;
      }
    }
  }
  for (int i = 0; i < x; i++) {
    for (int e = 0; e < y; e++) {
      if (table[i][e] == '*') {
        result += (row[i] - 1) * (col[e] - 1);
      }
    }
  }
  cout << result;
  return 0;
}

```