#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const int INF = (int)1e9;
const long double EPS = 1e-9;
const long double PI = 3.1415926535897932384626433832795;
int n, m;
char a[1010][1010];
long long sI[1010], sJ[1010];
int main() {
  cin >> n >> m;
  memset(sI, 0, sizeof(sI));
  memset(sJ, 0, sizeof(sJ));
  for (int i = 0; i < int(n); ++i)
    for (int j = 0; j < int(m); ++j) {
      scanf(" %c", &a[i][j]);
      if (a[i][j] == '*') {
        sI[i]++;
        sJ[j]++;
      }
    }
  long long ans = 0;
  for (int i = 0; i < int(n); ++i)
    for (int j = 0; j < int(m); ++j)
      if (a[i][j] == '*') ans += (sI[i] - 1) * (sJ[j] - 1);
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 comment(linker, "/STACK:256000000")
using namespace std;
const int INF = (int)1e9;
const long double EPS = 1e-9;
const long double PI = 3.1415926535897932384626433832795;
int n, m;
char a[1010][1010];
long long sI[1010], sJ[1010];
int main() {
  cin >> n >> m;
  memset(sI, 0, sizeof(sI));
  memset(sJ, 0, sizeof(sJ));
  for (int i = 0; i < int(n); ++i)
    for (int j = 0; j < int(m); ++j) {
      scanf(" %c", &a[i][j]);
      if (a[i][j] == '*') {
        sI[i]++;
        sJ[j]++;
      }
    }
  long long ans = 0;
  for (int i = 0; i < int(n); ++i)
    for (int j = 0; j < int(m); ++j)
      if (a[i][j] == '*') ans += (sI[i] - 1) * (sJ[j] - 1);
  cout << ans << endl;
  return 0;
}

```