#include <bits/stdc++.h>
using namespace std;
char str[1003];
long long ara[1003][1003];
long long row[1003][1003];
long long cumu[1003][1003];
long long func(long long r, long long c) {
  for (long long i = 1; i <= r; i++) {
    long long sum = 0;
    for (long long j = c; j >= 1; j--) row[i][j] = sum, sum += ara[i][j];
  }
  long long res = 0;
  for (long long i = 1; i <= c; i++) {
    long long sum = 0;
    for (long long j = r; j >= 1; j--)
      sum += (ara[j][i] * row[j][i]),
          cumu[j][i] = (sum - row[j][i]) * ara[j][i], res += cumu[j][i];
  }
  return res;
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  long long res = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%s", str);
    for (int j = 1; j <= m; j++)
      if (str[j - 1] == '*')
        ara[i][j] = 1;
      else
        ara[i][j] = 0;
  }
  res += func(n, m);
  for (int i = 1, k = m; i <= (m / 2); i++, k--)
    for (int j = 1; j <= n; j++) swap(ara[j][i], ara[j][k]);
  res += func(n, m);
  for (int i = 1, k = m; i <= (m / 2); i++, k--)
    for (int j = 1; j <= n; j++) swap(ara[j][i], ara[j][k]);
  for (int i = 1, k = n; i <= (n / 2); i++, k--)
    for (int j = 1; j <= m; j++) swap(ara[i][j], ara[k][j]);
  res += func(n, m);
  for (int i = 1, k = m; i <= (m / 2); i++, k--)
    for (int j = 1; j <= n; j++) swap(ara[j][i], ara[j][k]);
  res += func(n, m);
  cout << res << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
char str[1003];
long long ara[1003][1003];
long long row[1003][1003];
long long cumu[1003][1003];
long long func(long long r, long long c) {
  for (long long i = 1; i <= r; i++) {
    long long sum = 0;
    for (long long j = c; j >= 1; j--) row[i][j] = sum, sum += ara[i][j];
  }
  long long res = 0;
  for (long long i = 1; i <= c; i++) {
    long long sum = 0;
    for (long long j = r; j >= 1; j--)
      sum += (ara[j][i] * row[j][i]),
          cumu[j][i] = (sum - row[j][i]) * ara[j][i], res += cumu[j][i];
  }
  return res;
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  long long res = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%s", str);
    for (int j = 1; j <= m; j++)
      if (str[j - 1] == '*')
        ara[i][j] = 1;
      else
        ara[i][j] = 0;
  }
  res += func(n, m);
  for (int i = 1, k = m; i <= (m / 2); i++, k--)
    for (int j = 1; j <= n; j++) swap(ara[j][i], ara[j][k]);
  res += func(n, m);
  for (int i = 1, k = m; i <= (m / 2); i++, k--)
    for (int j = 1; j <= n; j++) swap(ara[j][i], ara[j][k]);
  for (int i = 1, k = n; i <= (n / 2); i++, k--)
    for (int j = 1; j <= m; j++) swap(ara[i][j], ara[k][j]);
  res += func(n, m);
  for (int i = 1, k = m; i <= (m / 2); i++, k--)
    for (int j = 1; j <= n; j++) swap(ara[j][i], ara[j][k]);
  res += func(n, m);
  cout << res << endl;
  return 0;
}

```