#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
int main(int argc, char** argv) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  bool f[n][m];
  for (long long i = 0, max_i = n; i < max_i; ++i)
    for (long long j = 0, max_i = m; j < max_i; ++j) {
      char x;
      cin >> x;
      f[i][j] = x == '*';
    }
  vector<long long> a[2];
  a[0].resize(n);
  a[1].resize(m);
  for (long long i = 0, max_i = n; i < max_i; ++i) {
    a[0][i] = 0;
    for (long long j = 0, max_i = m; j < max_i; ++j) a[0][i] += f[i][j];
  }
  for (long long j = 0, max_i = m; j < max_i; ++j) {
    a[1][j] = 0;
    for (long long i = 0, max_i = n; i < max_i; ++i) a[1][j] += f[i][j];
  }
  long long ans = 0;
  for (long long i = 0, max_i = n; i < max_i; ++i)
    for (long long j = 0, max_i = m; j < max_i; ++j)
      ans += f[i][j] * (a[0][i] - 1) * (a[1][j] - 1);
  cout << ans;
  return 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>
#pragma GCC optimize("O3")
using namespace std;
int main(int argc, char** argv) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  bool f[n][m];
  for (long long i = 0, max_i = n; i < max_i; ++i)
    for (long long j = 0, max_i = m; j < max_i; ++j) {
      char x;
      cin >> x;
      f[i][j] = x == '*';
    }
  vector<long long> a[2];
  a[0].resize(n);
  a[1].resize(m);
  for (long long i = 0, max_i = n; i < max_i; ++i) {
    a[0][i] = 0;
    for (long long j = 0, max_i = m; j < max_i; ++j) a[0][i] += f[i][j];
  }
  for (long long j = 0, max_i = m; j < max_i; ++j) {
    a[1][j] = 0;
    for (long long i = 0, max_i = n; i < max_i; ++i) a[1][j] += f[i][j];
  }
  long long ans = 0;
  for (long long i = 0, max_i = n; i < max_i; ++i)
    for (long long j = 0, max_i = m; j < max_i; ++j)
      ans += f[i][j] * (a[0][i] - 1) * (a[1][j] - 1);
  cout << ans;
  return 0;
}

```