#include <bits/stdc++.h>
using namespace std;
const long long Max = 1000 + 10;
char a[Max][Max];
long long c[Max], b[Max];
int main() {
  long long m, n;
  cin >> n >> m;
  for (long long i = 1; i <= n; ++i)
    for (long long j = 1; j <= m; ++j) cin >> a[i][j];
  for (long long i = 1; i <= n; ++i)
    for (long long j = 1; j <= m; ++j)
      if (a[i][j] == '*') ++b[i];
  for (long long i = 1; i <= m; ++i)
    for (long long j = 1; j <= n; ++j)
      if (a[j][i] == '*') ++c[i];
  long long sum = 0;
  for (long long i = 1; i <= n; ++i)
    for (long long j = 1; j <= m; ++j)
      if (a[i][j] == '*') sum += (b[i] - 1) * (c[j] - 1);
  cout << sum << endl;
}

### 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 long long Max = 1000 + 10;
char a[Max][Max];
long long c[Max], b[Max];
int main() {
  long long m, n;
  cin >> n >> m;
  for (long long i = 1; i <= n; ++i)
    for (long long j = 1; j <= m; ++j) cin >> a[i][j];
  for (long long i = 1; i <= n; ++i)
    for (long long j = 1; j <= m; ++j)
      if (a[i][j] == '*') ++b[i];
  for (long long i = 1; i <= m; ++i)
    for (long long j = 1; j <= n; ++j)
      if (a[j][i] == '*') ++c[i];
  long long sum = 0;
  for (long long i = 1; i <= n; ++i)
    for (long long j = 1; j <= m; ++j)
      if (a[i][j] == '*') sum += (b[i] - 1) * (c[j] - 1);
  cout << sum << endl;
}

```