#include <bits/stdc++.h>
using namespace std;
template <class T>
T abs(T x) {
  return x > 0 ? x : (-x);
}
template <class T>
T sqr(T x) {
  return x * x;
}
const int maxn = 1010;
int n, m;
char a[maxn][maxn];
long long h[maxn];
long long v[maxn];
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; ++i) scanf("%s", a[i]);
  for (int i = 0; i < n; ++i) {
    h[i] = 0;
    for (int j = 0; j < m; ++j)
      if (a[i][j] == '*') ++h[i];
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j)
      if (a[i][j] == '*') v[j] += h[i] - 1;
  long long res = 0;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) {
      if (a[i][j] != '*') continue;
      res += v[j] - h[i] + 1;
    }
  cout << res << "\n";
  return 0;
}

### Prompt

Please create a solution in cpp 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 abs(T x) {
  return x > 0 ? x : (-x);
}
template <class T>
T sqr(T x) {
  return x * x;
}
const int maxn = 1010;
int n, m;
char a[maxn][maxn];
long long h[maxn];
long long v[maxn];
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; ++i) scanf("%s", a[i]);
  for (int i = 0; i < n; ++i) {
    h[i] = 0;
    for (int j = 0; j < m; ++j)
      if (a[i][j] == '*') ++h[i];
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j)
      if (a[i][j] == '*') v[j] += h[i] - 1;
  long long res = 0;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) {
      if (a[i][j] != '*') continue;
      res += v[j] - h[i] + 1;
    }
  cout << res << "\n";
  return 0;
}

```