#include <bits/stdc++.h>
using namespace std;
char a[1010][1010];
int cnt1[1010], cnt2[1010];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = (0); i < (n); ++i) scanf("%s", &a[i]);
  memset(cnt1, 0, sizeof(cnt1));
  memset(cnt2, 0, sizeof(cnt2));
  for (int i = (0); i < (n); ++i)
    for (int j = (0); j < (m); ++j)
      if (a[i][j] == '*') {
        cnt1[i]++;
        cnt2[j]++;
      }
  long long res = 0;
  for (int i = (0); i < (n); ++i)
    for (int j = (0); j < (m); ++j)
      if (a[i][j] == '*') {
        res += (cnt1[i] - 1) * 1ll * (cnt2[j] - 1);
      }
  cout << res << endl;
  return 0;
}

### Prompt

Your task is to create 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;
char a[1010][1010];
int cnt1[1010], cnt2[1010];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = (0); i < (n); ++i) scanf("%s", &a[i]);
  memset(cnt1, 0, sizeof(cnt1));
  memset(cnt2, 0, sizeof(cnt2));
  for (int i = (0); i < (n); ++i)
    for (int j = (0); j < (m); ++j)
      if (a[i][j] == '*') {
        cnt1[i]++;
        cnt2[j]++;
      }
  long long res = 0;
  for (int i = (0); i < (n); ++i)
    for (int j = (0); j < (m); ++j)
      if (a[i][j] == '*') {
        res += (cnt1[i] - 1) * 1ll * (cnt2[j] - 1);
      }
  cout << res << endl;
  return 0;
}

```