#include <bits/stdc++.h>
using namespace std;
int n, m;
long long ss[1010][1010];
long long as[1010][1010];
long long bs[1010][1010];
long long cs[1010][1010];
long long ds[1010][1010];
string str;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    cin >> str;
    for (int j = 0; j < m; ++j) {
      ss[i][j] = str[j] == '*' ? 1 : 0;
      as[i][j] = ss[i][j];
      bs[i][j] = ss[i][j];
      cs[i][j] = ss[i][j];
      ds[i][j] = ss[i][j];
    }
  }
  for (int i = n - 1; i >= 0; --i)
    for (int j = m - 1; j >= 0; --j) {
      if (j < m - 1) as[i][j] = as[i][j + 1] + ss[i][j];
      if (i < n - 1) bs[i][j] = bs[i + 1][j] + ss[i][j];
    }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) {
      if (j > 0) cs[i][j] = cs[i][j - 1] + ss[i][j];
      if (i > 0) ds[i][j] = ds[i - 1][j] + ss[i][j];
    }
  long long ans = 0;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j)
      if (ss[i][j] == 1) {
        long long a, b, c, d;
        a = (j < m - 1) ? as[i][j + 1] : 0;
        b = (i < n - 1) ? bs[i + 1][j] : 0;
        c = (j > 0) ? cs[i][j - 1] : 0;
        d = (i > 0) ? ds[i - 1][j] : 0;
        ans += a * b + b * c + c * d + d * a;
      }
  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>
using namespace std;
int n, m;
long long ss[1010][1010];
long long as[1010][1010];
long long bs[1010][1010];
long long cs[1010][1010];
long long ds[1010][1010];
string str;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    cin >> str;
    for (int j = 0; j < m; ++j) {
      ss[i][j] = str[j] == '*' ? 1 : 0;
      as[i][j] = ss[i][j];
      bs[i][j] = ss[i][j];
      cs[i][j] = ss[i][j];
      ds[i][j] = ss[i][j];
    }
  }
  for (int i = n - 1; i >= 0; --i)
    for (int j = m - 1; j >= 0; --j) {
      if (j < m - 1) as[i][j] = as[i][j + 1] + ss[i][j];
      if (i < n - 1) bs[i][j] = bs[i + 1][j] + ss[i][j];
    }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) {
      if (j > 0) cs[i][j] = cs[i][j - 1] + ss[i][j];
      if (i > 0) ds[i][j] = ds[i - 1][j] + ss[i][j];
    }
  long long ans = 0;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j)
      if (ss[i][j] == 1) {
        long long a, b, c, d;
        a = (j < m - 1) ? as[i][j + 1] : 0;
        b = (i < n - 1) ? bs[i + 1][j] : 0;
        c = (j > 0) ? cs[i][j - 1] : 0;
        d = (i > 0) ? ds[i - 1][j] : 0;
        ans += a * b + b * c + c * d + d * a;
      }
  cout << ans << endl;
  return 0;
}

```