#include <bits/stdc++.h>
using namespace std;
int n, m;
char a[1100][1100];
int precalcI1[1100][1100];
int precalcI2[1100][1100];
int precalcJ1[1100][1100];
int precalcJ2[1100][1100];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < (int)n; ++i) scanf("%s", a[i]);
  memset(precalcI1, 0, sizeof(precalcI1));
  memset(precalcI2, 0, sizeof(precalcI2));
  memset(precalcJ1, 0, sizeof(precalcJ1));
  memset(precalcJ2, 0, sizeof(precalcJ2));
  for (int i = 0; i < (int)n; ++i) {
    for (int j = 1; j <= (int)m - 1; ++j) {
      precalcJ1[i][j] = precalcJ1[i][j - 1];
      if (a[i][j - 1] == '*') ++precalcJ1[i][j];
    }
    for (int j = m - 2; j >= 0; --j) {
      precalcJ2[i][j] = precalcJ2[i][j + 1];
      if (a[i][j + 1] == '*') ++precalcJ2[i][j];
    }
  }
  for (int j = 0; j < (int)m; ++j) {
    for (int i = 1; i <= (int)n - 1; ++i) {
      precalcI1[i][j] = precalcI1[i - 1][j];
      if (a[i - 1][j] == '*') ++precalcI1[i][j];
    }
    for (int i = n - 2; i >= 0; --i) {
      precalcI2[i][j] = precalcI2[i + 1][j];
      if (a[i + 1][j] == '*') ++precalcI2[i][j];
    }
  }
  long long result = 0;
  for (int i = 0; i < (int)n; ++i)
    for (int j = 0; j < (int)m; ++j)
      if (a[i][j] == '*') {
        result += (long long)precalcI1[i][j] * (long long)precalcJ1[i][j];
        result += (long long)precalcI2[i][j] * (long long)precalcJ2[i][j];
        result += (long long)precalcI1[i][j] * (long long)precalcJ2[i][j];
        result += (long long)precalcI2[i][j] * (long long)precalcJ1[i][j];
      }
  cout << result << endl;
  return 0;
}

### Prompt

Generate a CPP solution to 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;
int n, m;
char a[1100][1100];
int precalcI1[1100][1100];
int precalcI2[1100][1100];
int precalcJ1[1100][1100];
int precalcJ2[1100][1100];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < (int)n; ++i) scanf("%s", a[i]);
  memset(precalcI1, 0, sizeof(precalcI1));
  memset(precalcI2, 0, sizeof(precalcI2));
  memset(precalcJ1, 0, sizeof(precalcJ1));
  memset(precalcJ2, 0, sizeof(precalcJ2));
  for (int i = 0; i < (int)n; ++i) {
    for (int j = 1; j <= (int)m - 1; ++j) {
      precalcJ1[i][j] = precalcJ1[i][j - 1];
      if (a[i][j - 1] == '*') ++precalcJ1[i][j];
    }
    for (int j = m - 2; j >= 0; --j) {
      precalcJ2[i][j] = precalcJ2[i][j + 1];
      if (a[i][j + 1] == '*') ++precalcJ2[i][j];
    }
  }
  for (int j = 0; j < (int)m; ++j) {
    for (int i = 1; i <= (int)n - 1; ++i) {
      precalcI1[i][j] = precalcI1[i - 1][j];
      if (a[i - 1][j] == '*') ++precalcI1[i][j];
    }
    for (int i = n - 2; i >= 0; --i) {
      precalcI2[i][j] = precalcI2[i + 1][j];
      if (a[i + 1][j] == '*') ++precalcI2[i][j];
    }
  }
  long long result = 0;
  for (int i = 0; i < (int)n; ++i)
    for (int j = 0; j < (int)m; ++j)
      if (a[i][j] == '*') {
        result += (long long)precalcI1[i][j] * (long long)precalcJ1[i][j];
        result += (long long)precalcI2[i][j] * (long long)precalcJ2[i][j];
        result += (long long)precalcI1[i][j] * (long long)precalcJ2[i][j];
        result += (long long)precalcI2[i][j] * (long long)precalcJ1[i][j];
      }
  cout << result << endl;
  return 0;
}

```