#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:300000000")
using namespace std;
const double pi = 3.1415926535897932384626433832795;
template <class T>
T abs(T &a) {
  return a >= 0 ? a : -a;
};
template <class T>
T sqr(T &x) {
  return x * x;
}
vector<string> v;
int sumHor[2000][2000], sumVert[2000][2000];
int n, m;
void transpose() {
  vector<string> newV(m, string(n, ' '));
  for (int i = 0; i < (n); i++)
    for (int j = 0; j < (m); j++) newV[j][i] = v[i][m - j - 1];
  v = newV;
  swap(n, m);
}
int main() {
  cin >> n >> m;
  v.resize(n);
  for (int i = 0; i < (n); i++) cin >> v[i];
  long long ans = 0;
  for (int _ = 0; _ < (4); _++) {
    memset(sumHor, 0, sizeof(sumHor));
    memset(sumVert, 0, sizeof(sumVert));
    for (int i = 0; i < (n); i++)
      for (int j = 0; j < (m); j++) {
        int t = (v[i][j] == '*');
        if (i == 0)
          sumVert[i][j] = t;
        else
          sumVert[i][j] = sumVert[i - 1][j] + t;
        if (j == 0)
          sumHor[i][j] = t;
        else
          sumHor[i][j] = sumHor[i][j - 1] + t;
      }
    for (int i = 1; i < n; i++)
      for (int j = 1; j < m; j++)
        if (v[i][j] == '*') ans += sumVert[i - 1][j] * sumHor[i][j - 1];
    transpose();
  }
  cout << ans;
}

### 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>
#pragma comment(linker, "/STACK:300000000")
using namespace std;
const double pi = 3.1415926535897932384626433832795;
template <class T>
T abs(T &a) {
  return a >= 0 ? a : -a;
};
template <class T>
T sqr(T &x) {
  return x * x;
}
vector<string> v;
int sumHor[2000][2000], sumVert[2000][2000];
int n, m;
void transpose() {
  vector<string> newV(m, string(n, ' '));
  for (int i = 0; i < (n); i++)
    for (int j = 0; j < (m); j++) newV[j][i] = v[i][m - j - 1];
  v = newV;
  swap(n, m);
}
int main() {
  cin >> n >> m;
  v.resize(n);
  for (int i = 0; i < (n); i++) cin >> v[i];
  long long ans = 0;
  for (int _ = 0; _ < (4); _++) {
    memset(sumHor, 0, sizeof(sumHor));
    memset(sumVert, 0, sizeof(sumVert));
    for (int i = 0; i < (n); i++)
      for (int j = 0; j < (m); j++) {
        int t = (v[i][j] == '*');
        if (i == 0)
          sumVert[i][j] = t;
        else
          sumVert[i][j] = sumVert[i - 1][j] + t;
        if (j == 0)
          sumHor[i][j] = t;
        else
          sumHor[i][j] = sumHor[i][j - 1] + t;
      }
    for (int i = 1; i < n; i++)
      for (int j = 1; j < m; j++)
        if (v[i][j] == '*') ans += sumVert[i - 1][j] * sumHor[i][j - 1];
    transpose();
  }
  cout << ans;
}

```