#include <bits/stdc++.h>
using namespace std;
class loc {
 public:
  int x;
  int y;
  loc() {}
  loc(int a, int b) : x(a), y(b) {}
};
int main() {
  int n, m;
  cin >> n >> m;
  bool** arr = new bool*[n];
  queue<loc> rec;
  int ka = 0;
  for (int i = 0; i < n; i++) {
    arr[i] = new bool[m];
    for (int j = 0; j < m; j++) {
      char tmp;
      cin >> tmp;
      if (tmp == '*') {
        arr[i][j] = 1;
        rec.push(loc(i, j));
        ka++;
      } else
        arr[i][j] = 0;
    }
  }
  long long int ans = 0;
  long long int* dpx = new long long int[n];
  long long int* dpy = new long long int[m];
  fill(dpx, dpx + n, -1);
  fill(dpy, dpy + m, -1);
  while (!rec.empty()) {
    loc a = rec.front();
    rec.pop();
    long long int x = 0;
    long long int y = 0;
    if (dpy[a.y] == -1) {
      for (int i = 0; i < n; i++) {
        if (arr[i][a.y] == 1 && i != a.x) {
          y++;
        }
      }
      dpy[a.y] = y;
    } else {
      y = dpy[a.y];
    }
    if (dpx[a.x] == -1) {
      for (int i = 0; i < m; i++) {
        if (arr[a.x][i] == 1 && i != a.y) x++;
      }
      dpx[a.x] = x;
    } else {
      x = dpx[a.x];
    }
    ans += x * y;
  }
  cout << ans << endl;
  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>
using namespace std;
class loc {
 public:
  int x;
  int y;
  loc() {}
  loc(int a, int b) : x(a), y(b) {}
};
int main() {
  int n, m;
  cin >> n >> m;
  bool** arr = new bool*[n];
  queue<loc> rec;
  int ka = 0;
  for (int i = 0; i < n; i++) {
    arr[i] = new bool[m];
    for (int j = 0; j < m; j++) {
      char tmp;
      cin >> tmp;
      if (tmp == '*') {
        arr[i][j] = 1;
        rec.push(loc(i, j));
        ka++;
      } else
        arr[i][j] = 0;
    }
  }
  long long int ans = 0;
  long long int* dpx = new long long int[n];
  long long int* dpy = new long long int[m];
  fill(dpx, dpx + n, -1);
  fill(dpy, dpy + m, -1);
  while (!rec.empty()) {
    loc a = rec.front();
    rec.pop();
    long long int x = 0;
    long long int y = 0;
    if (dpy[a.y] == -1) {
      for (int i = 0; i < n; i++) {
        if (arr[i][a.y] == 1 && i != a.x) {
          y++;
        }
      }
      dpy[a.y] = y;
    } else {
      y = dpy[a.y];
    }
    if (dpx[a.x] == -1) {
      for (int i = 0; i < m; i++) {
        if (arr[a.x][i] == 1 && i != a.y) x++;
      }
      dpx[a.x] = x;
    } else {
      x = dpx[a.x];
    }
    ans += x * y;
  }
  cout << ans << endl;
  return 0;
}

```