#include <bits/stdc++.h>
using namespace std;
char s[1003][1003];
long long m, n, h[1003], w[1003], sum = 0;
signed main() {
  scanf("%d %d", &n, &m);
  for (long long i = 0; i < n; i++) {
    scanf("\n");
    for (long long j = 0; j < m; j++) scanf("%c", &s[i][j]);
  }
  for (long long i = 0; i < n; i++) h[i] = 0;
  for (long long j = 0; j < m; j++) w[j] = 0;
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < m; j++)
      if (s[i][j] == '*') {
        h[i]++;
        w[j]++;
      }
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < m; j++)
      if (s[i][j] == '*') sum += (h[i] - 1) * (w[j] - 1);
  printf("%lld", sum);
}

### 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 s[1003][1003];
long long m, n, h[1003], w[1003], sum = 0;
signed main() {
  scanf("%d %d", &n, &m);
  for (long long i = 0; i < n; i++) {
    scanf("\n");
    for (long long j = 0; j < m; j++) scanf("%c", &s[i][j]);
  }
  for (long long i = 0; i < n; i++) h[i] = 0;
  for (long long j = 0; j < m; j++) w[j] = 0;
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < m; j++)
      if (s[i][j] == '*') {
        h[i]++;
        w[j]++;
      }
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < m; j++)
      if (s[i][j] == '*') sum += (h[i] - 1) * (w[j] - 1);
  printf("%lld", sum);
}

```