#include <bits/stdc++.h>
int up[1010][1010];
char s[1010][1010];
int ll[1010][1010];
int lu[1010][1010];
int ru[1010][1010];
int main() {
  int n, m;
  while (scanf("%d%d", &n, &m) == 2) {
    for (int i = 1; i <= n; i++) scanf("%s", s[i] + 1);
    memset(up, 0, sizeof(up));
    memset(ll, 0, sizeof(ll));
    memset(lu, 0, sizeof(lu));
    memset(ru, 0, sizeof(ru));
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++) {
        up[i][j] = up[i - 1][j] + (s[i][j] == '*');
        ll[i][j] = ll[i][j - 1] + (s[i][j] == '*');
        lu[i][j] = lu[i - 1][j - 1] + (s[i][j] == '*');
        ru[i][j] = ru[i - 1][j + 1] + (s[i][j] == '*');
      }
    long long ans = 0;
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++) {
        if (s[i][j] != '*') continue;
        int UP = up[i][j];
        int DOWN = up[n][j] - UP;
        UP--;
        int left = ll[i][j];
        int right = ll[i][m] - left;
        left--;
        int LU = lu[i][j];
        int RU = ru[i][j];
        int ni, nj;
        if (i + j - 1 > n || i + j - 1 < 1) {
          ni = n;
          nj = i + j - n;
        } else {
          nj = 1;
          ni = i + j - 1;
        }
        int LD = ru[ni][nj] - RU;
        RU--;
        if (i + (m - j) > n || i + (m - j) < 1) {
          ni = n;
          nj = j + (n - i);
        } else {
          ni = i + (m - j);
          nj = m;
        }
        int RD = lu[ni][nj] - LU;
        LU--;
        ans += UP * left;
        ans += UP * right;
        ans += DOWN * left;
        ans += DOWN * right;
      }
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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>
int up[1010][1010];
char s[1010][1010];
int ll[1010][1010];
int lu[1010][1010];
int ru[1010][1010];
int main() {
  int n, m;
  while (scanf("%d%d", &n, &m) == 2) {
    for (int i = 1; i <= n; i++) scanf("%s", s[i] + 1);
    memset(up, 0, sizeof(up));
    memset(ll, 0, sizeof(ll));
    memset(lu, 0, sizeof(lu));
    memset(ru, 0, sizeof(ru));
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++) {
        up[i][j] = up[i - 1][j] + (s[i][j] == '*');
        ll[i][j] = ll[i][j - 1] + (s[i][j] == '*');
        lu[i][j] = lu[i - 1][j - 1] + (s[i][j] == '*');
        ru[i][j] = ru[i - 1][j + 1] + (s[i][j] == '*');
      }
    long long ans = 0;
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++) {
        if (s[i][j] != '*') continue;
        int UP = up[i][j];
        int DOWN = up[n][j] - UP;
        UP--;
        int left = ll[i][j];
        int right = ll[i][m] - left;
        left--;
        int LU = lu[i][j];
        int RU = ru[i][j];
        int ni, nj;
        if (i + j - 1 > n || i + j - 1 < 1) {
          ni = n;
          nj = i + j - n;
        } else {
          nj = 1;
          ni = i + j - 1;
        }
        int LD = ru[ni][nj] - RU;
        RU--;
        if (i + (m - j) > n || i + (m - j) < 1) {
          ni = n;
          nj = j + (n - i);
        } else {
          ni = i + (m - j);
          nj = m;
        }
        int RD = lu[ni][nj] - LU;
        LU--;
        ans += UP * left;
        ans += UP * right;
        ans += DOWN * left;
        ans += DOWN * right;
      }
    printf("%I64d\n", ans);
  }
  return 0;
}

```