#include <bits/stdc++.h>
using namespace std;
int main(int argc, char **argv) {
  ifstream from;
  const long long nmin = 1, nmax = 1000;
  long long n, m, rt, res, i, j, k, ax;
  char ch;
  if (0) from.open("test.txt");
  cin >> n >> m;
  vector<vector<char> > v(n);
  vector<long long> vx(n), vy(m);
  for (i = 0; i < n; i++) v[i].resize(m);
  for (i = 0; i < n; i++) vx[i] = 0;
  for (j = 0; j < m; j++) vy[j] = 0;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      cin >> ch;
      if (ch == '*') {
        v[i][j] = 1;
        vx[i]++;
        vy[j]++;
      } else
        v[i][j] = 0;
    }
  }
  res = 0;
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++)
      if (v[i][j] == 1) res += (vx[i] - 1) * (vy[j] - 1);
  cout << res << endl;
  return 0;
}

### Prompt

Generate 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;
int main(int argc, char **argv) {
  ifstream from;
  const long long nmin = 1, nmax = 1000;
  long long n, m, rt, res, i, j, k, ax;
  char ch;
  if (0) from.open("test.txt");
  cin >> n >> m;
  vector<vector<char> > v(n);
  vector<long long> vx(n), vy(m);
  for (i = 0; i < n; i++) v[i].resize(m);
  for (i = 0; i < n; i++) vx[i] = 0;
  for (j = 0; j < m; j++) vy[j] = 0;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      cin >> ch;
      if (ch == '*') {
        v[i][j] = 1;
        vx[i]++;
        vy[j]++;
      } else
        v[i][j] = 0;
    }
  }
  res = 0;
  for (i = 0; i < n; i++)
    for (j = 0; j < m; j++)
      if (v[i][j] == 1) res += (vx[i] - 1) * (vy[j] - 1);
  cout << res << endl;
  return 0;
}

```