#include <bits/stdc++.h>
using namespace std;
int N, M;
char m[1002][1002];
int s[1002];
long long sum;
int main(int argc, char *argv[]) {
  int i, j, tmp, tmp2;
  scanf("%d%d", &N, &M);
  for (i = 0; i < N; i++) scanf("%s", m[i]);
  sum = 0;
  for (i = 0; i < N; i++)
    for (j = 0; j < M; j++) {
      if (m[i][j] == '*') s[j]++;
    }
  for (i = 0; i < N; i++) {
    tmp = 0;
    tmp2 = 0;
    for (j = 0; j < M; j++) {
      if (m[i][j] == '*') {
        tmp++;
        tmp2 += s[j] - 1;
      }
    }
    sum += (long long)tmp2 * (tmp - 1);
  }
  printf("%lld\n", sum);
  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;
int N, M;
char m[1002][1002];
int s[1002];
long long sum;
int main(int argc, char *argv[]) {
  int i, j, tmp, tmp2;
  scanf("%d%d", &N, &M);
  for (i = 0; i < N; i++) scanf("%s", m[i]);
  sum = 0;
  for (i = 0; i < N; i++)
    for (j = 0; j < M; j++) {
      if (m[i][j] == '*') s[j]++;
    }
  for (i = 0; i < N; i++) {
    tmp = 0;
    tmp2 = 0;
    for (j = 0; j < M; j++) {
      if (m[i][j] == '*') {
        tmp++;
        tmp2 += s[j] - 1;
      }
    }
    sum += (long long)tmp2 * (tmp - 1);
  }
  printf("%lld\n", sum);
  return 0;
}

```