#include <bits/stdc++.h>
using namespace std;
template <class _T>
_T MAX(_T a, _T b) {
  return a > b ? a : b;
}
template <class _T>
_T MIN(_T a, _T b) {
  return a < b ? a : b;
}
const int N = 1024;
char board[N][N];
int CC[N], CL[N];
int main(void) {
  long long c = 0;
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < MAX(n, m); ++i) CL[i] = CC[i] = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%s", board[i]);
    for (int j = 0; j < m; ++j)
      if (board[i][j] == '*') ++CL[i], ++CC[j];
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j)
      if (board[i][j] == '*') c += 1LL * (CL[i] - 1) * (CC[j] - 1);
  printf("%lld\n", c);
  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>
using namespace std;
template <class _T>
_T MAX(_T a, _T b) {
  return a > b ? a : b;
}
template <class _T>
_T MIN(_T a, _T b) {
  return a < b ? a : b;
}
const int N = 1024;
char board[N][N];
int CC[N], CL[N];
int main(void) {
  long long c = 0;
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < MAX(n, m); ++i) CL[i] = CC[i] = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%s", board[i]);
    for (int j = 0; j < m; ++j)
      if (board[i][j] == '*') ++CL[i], ++CC[j];
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j)
      if (board[i][j] == '*') c += 1LL * (CL[i] - 1) * (CC[j] - 1);
  printf("%lld\n", c);
  return 0;
}

```