#include <bits/stdc++.h>
using namespace std;
const int N = 20010, K = 5001;
int t, n, m;
long long int f[2 * N], sol, x, x2, y, y2;
vector<pair<int, int> > hor[N], ver[N];
vector<int> endsInY[N];
void add(int x, long long int val) {
  for (; x < N; x += (x & -x)) f[x] += val;
  return;
}
long long int get(int x) {
  if (!x) return 0;
  long long int zbr = 0;
  for (; x > 0; x -= (x & -x)) zbr += f[x];
  return zbr;
}
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x >> y >> x2 >> y2;
    x += K;
    y += K;
    x2 += K;
    y2 += K;
    if (x == x2) ver[x].push_back(make_pair(min(y, y2), max(y, y2)));
    if (y == y2) hor[y].push_back(make_pair(min(x, x2), max(x, x2)));
  }
  for (int y = 1; y <= 10001; y++) {
    for (int currOnY = 0; currOnY < hor[y].size(); currOnY++) {
      for (int x = hor[y][currOnY].first; x <= hor[y][currOnY].second; x++) {
        for (int currOnX = 0; currOnX < ver[x].size(); currOnX++) {
          if (ver[x][currOnX].first <= y && y < ver[x][currOnX].second) {
            add(x, 1);
            endsInY[ver[x][currOnX].second].push_back(x);
          }
        }
      }
      for (int y2 = y + 1; y2 <= 10001; y2++) {
        for (int currOnY2 = 0; currOnY2 < hor[y2].size(); currOnY2++) {
          long long int addVal =
              get(hor[y2][currOnY2].second) - get(hor[y2][currOnY2].first - 1);
          sol += (addVal * (addVal - 1) / 2);
        }
        for (int i = 0; i < endsInY[y2].size(); i++) {
          add(endsInY[y2][i], -1);
        }
        endsInY[y2].clear();
      }
    }
  }
  cout << sol << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

There are n segments drawn on a plane; the i-th segment connects two points (x_{i, 1}, y_{i, 1}) and (x_{i, 2}, y_{i, 2}). Each segment is non-degenerate, and is either horizontal or vertical — formally, for every i ∈ [1, n] either x_{i, 1} = x_{i, 2} or y_{i, 1} = y_{i, 2} (but only one of these conditions holds). Only segments of different types may intersect: no pair of horizontal segments shares any common points, and no pair of vertical segments shares any common points.

We say that four segments having indices h_1, h_2, v_1 and v_2 such that h_1 < h_2 and v_1 < v_2 form a rectangle if the following conditions hold:

  * segments h_1 and h_2 are horizontal; 
  * segments v_1 and v_2 are vertical; 
  * segment h_1 intersects with segment v_1; 
  * segment h_2 intersects with segment v_1; 
  * segment h_1 intersects with segment v_2; 
  * segment h_2 intersects with segment v_2. 



Please calculate the number of ways to choose four segments so they form a rectangle. Note that the conditions h_1 < h_2 and v_1 < v_2 should hold.

Input

The first line contains one integer n (1 ≤ n ≤ 5000) — the number of segments.

Then n lines follow. The i-th line contains four integers x_{i, 1}, y_{i, 1}, x_{i, 2} and y_{i, 2} denoting the endpoints of the i-th segment. All coordinates of the endpoints are in the range [-5000, 5000].

It is guaranteed that each segment is non-degenerate and is either horizontal or vertical. Furthermore, if two segments share a common point, one of these segments is horizontal, and another one is vertical.

Output

Print one integer — the number of ways to choose four segments so they form a rectangle.

Examples

Input


7
-1 4 -1 -2
6 -1 -2 -1
-2 3 6 3
2 -2 2 4
4 -1 4 3
5 3 5 1
5 2 1 2


Output


7


Input


5
1 5 1 0
0 1 5 1
5 4 0 4
4 2 4 0
4 3 4 5


Output


0

Note

The following pictures represent sample cases:

<image> <image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 20010, K = 5001;
int t, n, m;
long long int f[2 * N], sol, x, x2, y, y2;
vector<pair<int, int> > hor[N], ver[N];
vector<int> endsInY[N];
void add(int x, long long int val) {
  for (; x < N; x += (x & -x)) f[x] += val;
  return;
}
long long int get(int x) {
  if (!x) return 0;
  long long int zbr = 0;
  for (; x > 0; x -= (x & -x)) zbr += f[x];
  return zbr;
}
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x >> y >> x2 >> y2;
    x += K;
    y += K;
    x2 += K;
    y2 += K;
    if (x == x2) ver[x].push_back(make_pair(min(y, y2), max(y, y2)));
    if (y == y2) hor[y].push_back(make_pair(min(x, x2), max(x, x2)));
  }
  for (int y = 1; y <= 10001; y++) {
    for (int currOnY = 0; currOnY < hor[y].size(); currOnY++) {
      for (int x = hor[y][currOnY].first; x <= hor[y][currOnY].second; x++) {
        for (int currOnX = 0; currOnX < ver[x].size(); currOnX++) {
          if (ver[x][currOnX].first <= y && y < ver[x][currOnX].second) {
            add(x, 1);
            endsInY[ver[x][currOnX].second].push_back(x);
          }
        }
      }
      for (int y2 = y + 1; y2 <= 10001; y2++) {
        for (int currOnY2 = 0; currOnY2 < hor[y2].size(); currOnY2++) {
          long long int addVal =
              get(hor[y2][currOnY2].second) - get(hor[y2][currOnY2].first - 1);
          sol += (addVal * (addVal - 1) / 2);
        }
        for (int i = 0; i < endsInY[y2].size(); i++) {
          add(endsInY[y2][i], -1);
        }
        endsInY[y2].clear();
      }
    }
  }
  cout << sol << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}

```