#include <bits/stdc++.h>
#pragma GCC optimize("Ofast", "unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
bool check(int x, pair<int, int> a) { return a.first <= x && x <= a.second; }
const int p = 32;
const int l = 5;
int sz;
struct bit {
  int *a;
  bit() {
    a = new int[sz];
    for (int i = 0; i < sz; ++i) a[i] = 0;
  }
  void set(int i) {
    int j = i >> l;
    a[j] |= 1 << (i - (j << l));
  }
};
int count(const bit &a, const bit &b) {
  int ans = 0;
  for (int i = 0; i < sz; ++i) ans += __builtin_popcount(a.a[i] & b.a[i]);
  return ans;
}
signed main() {
  int n;
  scanf("%d", &n);
  vector<pair<int, pair<int, int>>> vert;
  vector<pair<int, pair<int, int>>> hor;
  while (n--) {
    int x1, y1, x2, y2;
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2)
      vert.push_back({x1, {y1, y2}});
    else
      hor.push_back({y1, {x1, x2}});
  }
  if (signed(hor.size()) < signed(vert.size())) swap(hor, vert);
  sz = (signed(hor.size()) + p - 1) / p;
  vector<bit> inter(signed(vert.size()));
  for (int i = 0; i < signed(vert.size()); ++i)
    for (int j = 0; j < signed(hor.size()); ++j)
      if (check(hor[j].first, vert[i].second) &&
          check(vert[i].first, hor[j].second))
        inter[i].set(j);
  long long ans = 0;
  for (int i = 0; i < signed(vert.size()); ++i)
    for (int j = i + 1; j < signed(vert.size()); ++j) {
      int cnt = count(inter[i], inter[j]);
      ans += cnt * (cnt - 1) / 2;
    }
  printf("%I64d", ans);
  return 0;
}

### Prompt

Your challenge is to write 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>
#pragma GCC optimize("Ofast", "unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
bool check(int x, pair<int, int> a) { return a.first <= x && x <= a.second; }
const int p = 32;
const int l = 5;
int sz;
struct bit {
  int *a;
  bit() {
    a = new int[sz];
    for (int i = 0; i < sz; ++i) a[i] = 0;
  }
  void set(int i) {
    int j = i >> l;
    a[j] |= 1 << (i - (j << l));
  }
};
int count(const bit &a, const bit &b) {
  int ans = 0;
  for (int i = 0; i < sz; ++i) ans += __builtin_popcount(a.a[i] & b.a[i]);
  return ans;
}
signed main() {
  int n;
  scanf("%d", &n);
  vector<pair<int, pair<int, int>>> vert;
  vector<pair<int, pair<int, int>>> hor;
  while (n--) {
    int x1, y1, x2, y2;
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2)
      vert.push_back({x1, {y1, y2}});
    else
      hor.push_back({y1, {x1, x2}});
  }
  if (signed(hor.size()) < signed(vert.size())) swap(hor, vert);
  sz = (signed(hor.size()) + p - 1) / p;
  vector<bit> inter(signed(vert.size()));
  for (int i = 0; i < signed(vert.size()); ++i)
    for (int j = 0; j < signed(hor.size()); ++j)
      if (check(hor[j].first, vert[i].second) &&
          check(vert[i].first, hor[j].second))
        inter[i].set(j);
  long long ans = 0;
  for (int i = 0; i < signed(vert.size()); ++i)
    for (int j = i + 1; j < signed(vert.size()); ++j) {
      int cnt = count(inter[i], inter[j]);
      ans += cnt * (cnt - 1) / 2;
    }
  printf("%I64d", ans);
  return 0;
}

```