#include <bits/stdc++.h>
using namespace std;
const double EPS = 1e-8;
const int INF = 0x3F3F3F3F;
const int64_t LINF = 4611686018427387903;
const int MOD = 1e9 + 7;
struct seg {
  int x1, y1, x2, y2;
};
bool cmp(seg a, seg b) { return a.x1 < b.x1; }
const int N = 10025;
int bit[N];
void mod(int pos, int val) {
  for (; pos < N; pos += (pos & -pos)) bit[pos] += val;
}
int que(int pos) {
  int res = 0;
  for (; pos; pos -= (pos & -pos)) res += bit[pos];
  return res;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  int n;
  vector<seg> a, b;
  cin >> n;
  for (int i = 0, x, y, z, r; i < n; i++) {
    cin >> x >> y >> z >> r;
    x += 5001, y += 5001, z += 5001, r += 5001;
    if (x > z) swap(x, z);
    if (y > r) swap(y, r);
    if (x == z)
      a.push_back({x, y, z, r});
    else
      b.push_back({x, y, z, r});
  }
  sort(a.begin(), a.end(), cmp);
  sort(b.begin(), b.end(), cmp);
  int64_t ans = 0;
  for (int i = 1; i < a.size(); i++) {
    memset(bit, 0, sizeof bit);
    for (int j = 0, it = 0; j < i; j++) {
      while (it < b.size() && b[it].x1 <= a[j].x1) {
        if (b[it].x2 >= a[i].x1) mod(b[it].y1, 1);
        it++;
      }
      int l = max(a[i].y1, a[j].y1), r = min(a[i].y2, a[j].y2);
      if (l > r) continue;
      int64_t cnt = que(r) - que(l - 1);
      ans += cnt * (cnt - 1) / 2;
    }
  }
  cout << 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>
using namespace std;
const double EPS = 1e-8;
const int INF = 0x3F3F3F3F;
const int64_t LINF = 4611686018427387903;
const int MOD = 1e9 + 7;
struct seg {
  int x1, y1, x2, y2;
};
bool cmp(seg a, seg b) { return a.x1 < b.x1; }
const int N = 10025;
int bit[N];
void mod(int pos, int val) {
  for (; pos < N; pos += (pos & -pos)) bit[pos] += val;
}
int que(int pos) {
  int res = 0;
  for (; pos; pos -= (pos & -pos)) res += bit[pos];
  return res;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  int n;
  vector<seg> a, b;
  cin >> n;
  for (int i = 0, x, y, z, r; i < n; i++) {
    cin >> x >> y >> z >> r;
    x += 5001, y += 5001, z += 5001, r += 5001;
    if (x > z) swap(x, z);
    if (y > r) swap(y, r);
    if (x == z)
      a.push_back({x, y, z, r});
    else
      b.push_back({x, y, z, r});
  }
  sort(a.begin(), a.end(), cmp);
  sort(b.begin(), b.end(), cmp);
  int64_t ans = 0;
  for (int i = 1; i < a.size(); i++) {
    memset(bit, 0, sizeof bit);
    for (int j = 0, it = 0; j < i; j++) {
      while (it < b.size() && b[it].x1 <= a[j].x1) {
        if (b[it].x2 >= a[i].x1) mod(b[it].y1, 1);
        it++;
      }
      int l = max(a[i].y1, a[j].y1), r = min(a[i].y2, a[j].y2);
      if (l > r) continue;
      int64_t cnt = que(r) - que(l - 1);
      ans += cnt * (cnt - 1) / 2;
    }
  }
  cout << ans;
  return 0;
}

```