#include <bits/stdc++.h>
using namespace std;
long long row[1000], col[1000];
char ch[1000][1000];
list<int> ax, ay;
int main() {
  int i, j, x, y;
  long long res = 0;
  cin >> x >> y;
  memset((row), (-1), sizeof((row)));
  memset((col), (-1), sizeof((col)));
  for (i = 0; i < x; i++) {
    for (j = 0; j < y; j++) {
      cin >> ch[i][j];
      if (ch[i][j] == '*') {
        ax.push_back(i);
        ay.push_back(j);
        row[i]++;
        col[j]++;
      }
    }
  }
  list<int>::iterator it, it2;
  it = ax.begin();
  it2 = ay.begin();
  while (it != ax.end()) {
    if (row[*it] >= 1 && col[*it2] >= 1) res += row[*it] * col[*it2];
    it++;
    it2++;
  }
  cout << res << '\n';
  return 0;
}

### Prompt

Please formulate a CPP solution 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>
using namespace std;
long long row[1000], col[1000];
char ch[1000][1000];
list<int> ax, ay;
int main() {
  int i, j, x, y;
  long long res = 0;
  cin >> x >> y;
  memset((row), (-1), sizeof((row)));
  memset((col), (-1), sizeof((col)));
  for (i = 0; i < x; i++) {
    for (j = 0; j < y; j++) {
      cin >> ch[i][j];
      if (ch[i][j] == '*') {
        ax.push_back(i);
        ay.push_back(j);
        row[i]++;
        col[j]++;
      }
    }
  }
  list<int>::iterator it, it2;
  it = ax.begin();
  it2 = ay.begin();
  while (it != ax.end()) {
    if (row[*it] >= 1 && col[*it2] >= 1) res += row[*it] * col[*it2];
    it++;
    it2++;
  }
  cout << res << '\n';
  return 0;
}

```