#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  char cel[1000][1000];
  cin.ignore();
  for (int i = 0; i < (int)(n); i++) {
    for (int j = 0; j < (int)(m); j++) scanf("%c", cel[i] + j);
    cin.ignore();
  }
  int hol[1000], ver[1000];
  for (int i = 0; i < (int)(n); i++) {
    int s = 0;
    for (int j = 0; j < (int)(m); j++) s += cel[i][j] == '*';
    hol[i] = s - 1;
  }
  for (int i = 0; i < (int)(m); i++) {
    int s = 0;
    for (int j = 0; j < (int)(n); j++) s += cel[j][i] == '*';
    ver[i] = s - 1;
  }
  long long ans = 0;
  for (int i = 0; i < (int)(n); i++)
    for (int j = 0; j < (int)(m); j++)
      if (cel[i][j] == '*') ans += hol[i] * ver[j];
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 main() {
  int n, m;
  scanf("%d%d", &n, &m);
  char cel[1000][1000];
  cin.ignore();
  for (int i = 0; i < (int)(n); i++) {
    for (int j = 0; j < (int)(m); j++) scanf("%c", cel[i] + j);
    cin.ignore();
  }
  int hol[1000], ver[1000];
  for (int i = 0; i < (int)(n); i++) {
    int s = 0;
    for (int j = 0; j < (int)(m); j++) s += cel[i][j] == '*';
    hol[i] = s - 1;
  }
  for (int i = 0; i < (int)(m); i++) {
    int s = 0;
    for (int j = 0; j < (int)(n); j++) s += cel[j][i] == '*';
    ver[i] = s - 1;
  }
  long long ans = 0;
  for (int i = 0; i < (int)(n); i++)
    for (int j = 0; j < (int)(m); j++)
      if (cel[i][j] == '*') ans += hol[i] * ver[j];
  printf("%I64d\n", ans);
  return 0;
}

```