#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:100000000000,100000000000")
const long long INF = 1000000007;
const long double cp = 2 * asin(1.0);
const long double eps = 1e-9;
const long long mod = 1000000007;
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  char s[1001][1001];
  long long a[1001], b[1001], ans = 0;
  memset(a, sizeof(a), 0);
  memset(b, sizeof(b), 0);
  cin >> n >> m;
  for (int(i) = 0; (i) < (n); i++)
    for (int(j) = 0; (j) < (m); j++) {
      cin >> s[i][j];
      if (s[i][j] == '*') a[i]++, b[j]++;
    }
  for (int(i) = 0; (i) < (n); i++)
    for (int(j) = 0; (j) < (m); j++)
      if (s[i][j] == '*') ans += (a[i] - 1) * (b[j] - 1);
  cout << ans;
  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>
#pragma comment(linker, "/STACK:100000000000,100000000000")
const long long INF = 1000000007;
const long double cp = 2 * asin(1.0);
const long double eps = 1e-9;
const long long mod = 1000000007;
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  char s[1001][1001];
  long long a[1001], b[1001], ans = 0;
  memset(a, sizeof(a), 0);
  memset(b, sizeof(b), 0);
  cin >> n >> m;
  for (int(i) = 0; (i) < (n); i++)
    for (int(j) = 0; (j) < (m); j++) {
      cin >> s[i][j];
      if (s[i][j] == '*') a[i]++, b[j]++;
    }
  for (int(i) = 0; (i) < (n); i++)
    for (int(j) = 0; (j) < (m); j++)
      if (s[i][j] == '*') ans += (a[i] - 1) * (b[j] - 1);
  cout << ans;
  return 0;
}

```