#include <bits/stdc++.h>
using namespace std;
char mat[1100][1100];
int N, M;
int col[1100], row[1100];
int main() {
  scanf("%d %d", &N, &M);
  for (int i = 0; i < N; i++) {
    scanf("%s", mat[i]);
  }
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < M; j++) {
      if (mat[i][j] == '*') row[i]++;
    }
  }
  for (int i = 0; i < M; i++) {
    for (int j = 0; j < N; j++) {
      if (mat[j][i] == '*') col[i]++;
    }
  }
  long long res = 0;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < M; j++) {
      if ('*' == mat[i][j]) {
        res += (row[i] - 1) * (col[j] - 1);
      }
    }
  }
  printf("%I64d\n", res);
  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;
char mat[1100][1100];
int N, M;
int col[1100], row[1100];
int main() {
  scanf("%d %d", &N, &M);
  for (int i = 0; i < N; i++) {
    scanf("%s", mat[i]);
  }
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < M; j++) {
      if (mat[i][j] == '*') row[i]++;
    }
  }
  for (int i = 0; i < M; i++) {
    for (int j = 0; j < N; j++) {
      if (mat[j][i] == '*') col[i]++;
    }
  }
  long long res = 0;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < M; j++) {
      if ('*' == mat[i][j]) {
        res += (row[i] - 1) * (col[j] - 1);
      }
    }
  }
  printf("%I64d\n", res);
  return 0;
}

```