#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 30;
int n, m;
char ma[2000][2000];
int dp[1500];
int main() {
  int i, j, k;
  cin >> n >> m;
  for (i = 0; i < n; i++) {
    scanf("%s", &ma[i]);
  }
  for (i = 0; i < n; i++) {
    int cnt = 0;
    for (j = 0; j < m; j++) {
      if (ma[i][j] == '*') cnt++;
    }
    dp[i] = cnt;
  }
  long long ans = 0;
  for (i = 0; i < m; i++) {
    int cnt = 0, num = 0;
    for (j = 0; j < n; j++) {
      if (ma[j][i] == '*') {
        cnt++;
        num += (dp[j] - 1);
      }
    }
    if (cnt >= 2) {
      ans += (num * (cnt - 1));
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Create a solution in CPP for 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;
const int INF = 1 << 30;
int n, m;
char ma[2000][2000];
int dp[1500];
int main() {
  int i, j, k;
  cin >> n >> m;
  for (i = 0; i < n; i++) {
    scanf("%s", &ma[i]);
  }
  for (i = 0; i < n; i++) {
    int cnt = 0;
    for (j = 0; j < m; j++) {
      if (ma[i][j] == '*') cnt++;
    }
    dp[i] = cnt;
  }
  long long ans = 0;
  for (i = 0; i < m; i++) {
    int cnt = 0, num = 0;
    for (j = 0; j < n; j++) {
      if (ma[j][i] == '*') {
        cnt++;
        num += (dp[j] - 1);
      }
    }
    if (cnt >= 2) {
      ans += (num * (cnt - 1));
    }
  }
  printf("%I64d\n", ans);
  return 0;
}

```