#include <bits/stdc++.h>
char p[1005][1005];
long long int h[40];
int row[1050], column[1050];
struct edge {
  int x;
  int y;
} pos[1000002];
int main() {
  int len, i, j, l, a, b, n, m, k, count;
  long long int sum;
  while (scanf("%d%d", &n, &m) != EOF) {
    getchar();
    count = 0;
    sum = 0;
    memset(row, 0, sizeof(row));
    memset(column, 0, sizeof(column));
    for (i = 1; i <= n; i++)
      for (j = 1; j <= m; j++) {
        scanf("%c", &p[i][j]);
        if (j == m) getchar();
        if (p[i][j] == '*') {
          pos[count].x = i;
          pos[count].y = j;
          count++;
          row[i]++;
          column[j]++;
        }
      }
    for (i = 0; i < count; i++) {
      sum += (row[pos[i].x] - 1) * (column[pos[i].y] - 1);
    }
    printf("%lld\n", sum);
  }
}

### Prompt

Please create a solution in CPP 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>
char p[1005][1005];
long long int h[40];
int row[1050], column[1050];
struct edge {
  int x;
  int y;
} pos[1000002];
int main() {
  int len, i, j, l, a, b, n, m, k, count;
  long long int sum;
  while (scanf("%d%d", &n, &m) != EOF) {
    getchar();
    count = 0;
    sum = 0;
    memset(row, 0, sizeof(row));
    memset(column, 0, sizeof(column));
    for (i = 1; i <= n; i++)
      for (j = 1; j <= m; j++) {
        scanf("%c", &p[i][j]);
        if (j == m) getchar();
        if (p[i][j] == '*') {
          pos[count].x = i;
          pos[count].y = j;
          count++;
          row[i]++;
          column[j]++;
        }
      }
    for (i = 0; i < count; i++) {
      sum += (row[pos[i].x] - 1) * (column[pos[i].y] - 1);
    }
    printf("%lld\n", sum);
  }
}

```