#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
template <typename T1, typename T2>
inline void chkmin(T1 &x, const T2 &y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1 &x, const T2 &y) {
  if (x < y) x = y;
}
const ll MAXN = 1010;
char table[MAXN][MAXN];
ll n, m;
void read() {
  cin >> n >> m;
  for (ll i = 0; i < n; i++) {
    for (ll j = 0; j < m; j++) {
      cin >> table[i][j];
    }
  }
}
ll ans;
ll cnt[MAXN][2];
void build() {
  for (ll i = 0; i < n; i++) {
    for (ll j = 0; j < m; j++) {
      if (table[i][j] == '*') {
        cnt[i][0]++;
        cnt[j][1]++;
      }
    }
  }
}
void calc() {
  for (ll i = 0; i < n; i++) {
    for (ll j = 0; j < m; j++) {
      if (table[i][j] == '*') {
        ans += (cnt[i][0] - 1) * (cnt[j][1] - 1);
      }
    }
  }
}
void run() {
  ans = 0;
  build();
  calc();
}
void write() { cout << ans << endl; }
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  read();
  run();
  write();
  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;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
template <typename T1, typename T2>
inline void chkmin(T1 &x, const T2 &y) {
  if (x > y) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1 &x, const T2 &y) {
  if (x < y) x = y;
}
const ll MAXN = 1010;
char table[MAXN][MAXN];
ll n, m;
void read() {
  cin >> n >> m;
  for (ll i = 0; i < n; i++) {
    for (ll j = 0; j < m; j++) {
      cin >> table[i][j];
    }
  }
}
ll ans;
ll cnt[MAXN][2];
void build() {
  for (ll i = 0; i < n; i++) {
    for (ll j = 0; j < m; j++) {
      if (table[i][j] == '*') {
        cnt[i][0]++;
        cnt[j][1]++;
      }
    }
  }
}
void calc() {
  for (ll i = 0; i < n; i++) {
    for (ll j = 0; j < m; j++) {
      if (table[i][j] == '*') {
        ans += (cnt[i][0] - 1) * (cnt[j][1] - 1);
      }
    }
  }
}
void run() {
  ans = 0;
  build();
  calc();
}
void write() { cout << ans << endl; }
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  read();
  run();
  write();
  return 0;
}

```