#include <bits/stdc++.h>
using namespace std;
struct newint {
  long long x = 0, y = 0;
};
const int N = 1000 + 100;
newint a[N];
bool mark[N][N];
int main() {
  int beginT = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  newint d;
  char k;
  long long ans = 0;
  cin >> d.y >> d.x;
  for (int i = 0; i < d.y; ++i)
    for (int j = 0; j < d.x; ++j) {
      cin >> k;
      if (k == '*') {
        ++a[j].x;
        ++a[i].y;
        mark[i][j] = 1;
      }
    }
  for (int i = 0; i < d.y; ++i)
    for (int j = 0; j < d.x; ++j) {
      if (mark[i][j]) {
        ans += (a[i].y - 1) * (a[j].x - 1);
      }
    }
  cout << ans;
  return cerr << "Time: " << (1. * clock() - beginT) / CLOCKS_PER_SEC << endl,
         0;
}

### Prompt

Your challenge is to write 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;
struct newint {
  long long x = 0, y = 0;
};
const int N = 1000 + 100;
newint a[N];
bool mark[N][N];
int main() {
  int beginT = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  newint d;
  char k;
  long long ans = 0;
  cin >> d.y >> d.x;
  for (int i = 0; i < d.y; ++i)
    for (int j = 0; j < d.x; ++j) {
      cin >> k;
      if (k == '*') {
        ++a[j].x;
        ++a[i].y;
        mark[i][j] = 1;
      }
    }
  for (int i = 0; i < d.y; ++i)
    for (int j = 0; j < d.x; ++j) {
      if (mark[i][j]) {
        ans += (a[i].y - 1) * (a[j].x - 1);
      }
    }
  cout << ans;
  return cerr << "Time: " << (1. * clock() - beginT) / CLOCKS_PER_SEC << endl,
         0;
}

```