#include <bits/stdc++.h>
int n, m;
char s[1011][1011];
int r[1011][1011], c[1011][1011], rs[1011], cs[1011];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%s", s[i]);
    for (int j = m; j > 0; j--) s[i][j] = s[i][j - 1];
    for (int j = 1; j <= m; j++) {
      c[i][j] = c[i - 1][j];
      r[i][j] = r[i][j - 1];
      if (s[i][j] == '*') {
        rs[i]++;
        cs[j]++;
        r[i][j]++;
        c[i][j]++;
      }
    }
  }
  long long int ans = 0LL;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
      if (s[i][j] == '*') {
        int a = r[i][j - 1], b = c[i - 1][j];
        int c = rs[i] - a - 1, d = cs[j] - b - 1;
        ans += 1LL * a * b + 1LL * b * c + 1LL * c * d + 1LL * d * a;
      }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
int n, m;
char s[1011][1011];
int r[1011][1011], c[1011][1011], rs[1011], cs[1011];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%s", s[i]);
    for (int j = m; j > 0; j--) s[i][j] = s[i][j - 1];
    for (int j = 1; j <= m; j++) {
      c[i][j] = c[i - 1][j];
      r[i][j] = r[i][j - 1];
      if (s[i][j] == '*') {
        rs[i]++;
        cs[j]++;
        r[i][j]++;
        c[i][j]++;
      }
    }
  }
  long long int ans = 0LL;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
      if (s[i][j] == '*') {
        int a = r[i][j - 1], b = c[i - 1][j];
        int c = rs[i] - a - 1, d = cs[j] - b - 1;
        ans += 1LL * a * b + 1LL * b * c + 1LL * c * d + 1LL * d * a;
      }
  printf("%I64d\n", ans);
  return 0;
}

```