#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1010;
const int MAXM = 6000010;
const int INF = 0x3f3f3f3f;
const int mod = 1000003;
const double eps = 1e-8;
char mp[MAXN][MAXN];
int dp[4][MAXN][MAXN];
int main() {
  int n, m;
  while (scanf("%d%d", &n, &m) != EOF) {
    for (int i = 1; i <= n; i++) {
      getchar();
      for (int j = 1; j <= m; j++) scanf("%c", &mp[i][j]);
    }
    memset(dp, 0, sizeof(dp));
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        if (mp[i][j] == '*') {
          dp[0][i][j] = dp[0][i][j - 1] + 1;
        } else {
          dp[0][i][j] = dp[0][i][j - 1];
        }
      }
      for (int j = m; j >= 1; j--) {
        if (mp[i][j] == '*') {
          dp[1][i][j] = dp[1][i][j + 1] + 1;
        } else {
          dp[1][i][j] = dp[1][i][j + 1];
        }
      }
    }
    for (int i = 1; i <= m; i++) {
      for (int j = 1; j <= n; j++) {
        if (mp[j][i] == '*') {
          dp[2][j][i] = dp[2][j - 1][i] + 1;
        } else {
          dp[2][j][i] = dp[2][j - 1][i];
        }
      }
      for (int j = n; j >= 1; j--) {
        if (mp[j][i] == '*') {
          dp[3][j][i] = dp[3][j + 1][i] + 1;
        } else {
          dp[3][j][i] = dp[3][j + 1][i];
        }
      }
    }
    long long sum = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        if (mp[i][j] == '*') {
          sum += (dp[0][i][j] - 1) * (dp[2][i][j] - 1);
          sum += (dp[0][i][j] - 1) * (dp[3][i][j] - 1);
          sum += (dp[1][i][j] - 1) * (dp[2][i][j] - 1);
          sum += (dp[1][i][j] - 1) * (dp[3][i][j] - 1);
        }
      }
    }
    cout << sum << 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;
const int MAXN = 1010;
const int MAXM = 6000010;
const int INF = 0x3f3f3f3f;
const int mod = 1000003;
const double eps = 1e-8;
char mp[MAXN][MAXN];
int dp[4][MAXN][MAXN];
int main() {
  int n, m;
  while (scanf("%d%d", &n, &m) != EOF) {
    for (int i = 1; i <= n; i++) {
      getchar();
      for (int j = 1; j <= m; j++) scanf("%c", &mp[i][j]);
    }
    memset(dp, 0, sizeof(dp));
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        if (mp[i][j] == '*') {
          dp[0][i][j] = dp[0][i][j - 1] + 1;
        } else {
          dp[0][i][j] = dp[0][i][j - 1];
        }
      }
      for (int j = m; j >= 1; j--) {
        if (mp[i][j] == '*') {
          dp[1][i][j] = dp[1][i][j + 1] + 1;
        } else {
          dp[1][i][j] = dp[1][i][j + 1];
        }
      }
    }
    for (int i = 1; i <= m; i++) {
      for (int j = 1; j <= n; j++) {
        if (mp[j][i] == '*') {
          dp[2][j][i] = dp[2][j - 1][i] + 1;
        } else {
          dp[2][j][i] = dp[2][j - 1][i];
        }
      }
      for (int j = n; j >= 1; j--) {
        if (mp[j][i] == '*') {
          dp[3][j][i] = dp[3][j + 1][i] + 1;
        } else {
          dp[3][j][i] = dp[3][j + 1][i];
        }
      }
    }
    long long sum = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        if (mp[i][j] == '*') {
          sum += (dp[0][i][j] - 1) * (dp[2][i][j] - 1);
          sum += (dp[0][i][j] - 1) * (dp[3][i][j] - 1);
          sum += (dp[1][i][j] - 1) * (dp[2][i][j] - 1);
          sum += (dp[1][i][j] - 1) * (dp[3][i][j] - 1);
        }
      }
    }
    cout << sum << endl;
  }
  return 0;
}

```