#include <bits/stdc++.h>
using namespace std;
vector<int> col;
vector<string> s;
int main() {
  int R, C, cnt;
  long long res = 0;
  cin >> R >> C;
  s.assign(R, string(C, ' '));
  for (int r = 0; r < (R); ++r) cin >> s[r];
  col.assign(C, 0);
  for (int r = 0; r < (R); ++r) {
    cnt = 0;
    for (int c = 0; c < (C); ++c)
      if (s[r][c] == '*') {
        res += cnt * col[c];
        cnt++;
      }
    cnt = 0;
    for (int c = (C - 1); c >= (0); --c)
      if (s[r][c] == '*') {
        res += cnt * col[c];
        cnt++;
        col[c]++;
      }
  }
  col.assign(C, 0);
  for (int r = (R - 1); r >= (0); --r) {
    cnt = 0;
    for (int c = 0; c < (C); ++c)
      if (s[r][c] == '*') {
        res += cnt * col[c];
        cnt++;
      }
    cnt = 0;
    for (int c = (C - 1); c >= (0); --c)
      if (s[r][c] == '*') {
        res += cnt * col[c];
        cnt++;
        col[c]++;
      }
  }
  cout << res << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
vector<int> col;
vector<string> s;
int main() {
  int R, C, cnt;
  long long res = 0;
  cin >> R >> C;
  s.assign(R, string(C, ' '));
  for (int r = 0; r < (R); ++r) cin >> s[r];
  col.assign(C, 0);
  for (int r = 0; r < (R); ++r) {
    cnt = 0;
    for (int c = 0; c < (C); ++c)
      if (s[r][c] == '*') {
        res += cnt * col[c];
        cnt++;
      }
    cnt = 0;
    for (int c = (C - 1); c >= (0); --c)
      if (s[r][c] == '*') {
        res += cnt * col[c];
        cnt++;
        col[c]++;
      }
  }
  col.assign(C, 0);
  for (int r = (R - 1); r >= (0); --r) {
    cnt = 0;
    for (int c = 0; c < (C); ++c)
      if (s[r][c] == '*') {
        res += cnt * col[c];
        cnt++;
      }
    cnt = 0;
    for (int c = (C - 1); c >= (0); --c)
      if (s[r][c] == '*') {
        res += cnt * col[c];
        cnt++;
        col[c]++;
      }
  }
  cout << res << endl;
  return 0;
}

```