#include <bits/stdc++.h>
using namespace std;
const int maxN = 1e3 + 3;
long long n, m, SumC[maxN], SumR[maxN];
long long ans;
bool table[maxN][maxN];
char c;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      cin >> c;
      table[i][j] = (c == '*' ? 1 : 0);
    }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (table[i][j]) {
        SumC[i]++;
        SumR[j]++;
      }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      ans += (table[i][j] ? (SumC[i] - 1) * (SumR[j] - 1) : 0);
  cout << ans;
}

### 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;
const int maxN = 1e3 + 3;
long long n, m, SumC[maxN], SumR[maxN];
long long ans;
bool table[maxN][maxN];
char c;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      cin >> c;
      table[i][j] = (c == '*' ? 1 : 0);
    }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (table[i][j]) {
        SumC[i]++;
        SumR[j]++;
      }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      ans += (table[i][j] ? (SumC[i] - 1) * (SumR[j] - 1) : 0);
  cout << ans;
}

```