#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e4 + 10, D = 5e3 + 1;
int n;
vector<pair<int, int> > h[maxn], v[maxn];
vector<int> d[maxn];
long long ans, sum[maxn];
int lowbit(int x) { return x & (-x); }
void add(int qq, int val) {
  while (qq <= maxn) {
    sum[qq] += val;
    qq += lowbit(qq);
  }
}
int get(int qq) {
  long long res = 0;
  while (qq) {
    res += sum[qq];
    qq -= lowbit(qq);
  }
  return res;
}
int get(int L, int R) { return get(R) - get(L - 1); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x1, x2, y1, y2;
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    x1 += D;
    x2 += D;
    y1 += D;
    y2 += D;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2) {
      v[x1].push_back(make_pair(y1, y2));
    } else {
      h[y1].push_back(make_pair(x1, x2));
    }
  }
  for (int y = 0; y < maxn; y++)
    for (int i = 0; i < h[y].size(); i++) {
      pair<int, int> t = h[y][i];
      for (int x = 0; x < maxn; x++) d[x].clear();
      for (int x = t.first; x <= t.second; x++)
        for (int j = 0; j < v[x].size(); j++) {
          pair<int, int> t2 = v[x][j];
          if (t2.first <= y && t2.second > y) {
            add(x, 1);
            d[t2.second].push_back(x);
          }
        }
      for (int y2 = y + 1; y2 < maxn; y2++) {
        for (int j = 0; j < h[y2].size(); j++) {
          pair<int, int> t2 = h[y2][j];
          long long cur = get(t2.first, t2.second);
          ans += cur * (cur - 1) / 2;
        }
        for (int j = 0; j < d[y2].size(); j++) {
          add(d[y2][j], -1);
        }
      }
    }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 maxn = 1e4 + 10, D = 5e3 + 1;
int n;
vector<pair<int, int> > h[maxn], v[maxn];
vector<int> d[maxn];
long long ans, sum[maxn];
int lowbit(int x) { return x & (-x); }
void add(int qq, int val) {
  while (qq <= maxn) {
    sum[qq] += val;
    qq += lowbit(qq);
  }
}
int get(int qq) {
  long long res = 0;
  while (qq) {
    res += sum[qq];
    qq -= lowbit(qq);
  }
  return res;
}
int get(int L, int R) { return get(R) - get(L - 1); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x1, x2, y1, y2;
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    x1 += D;
    x2 += D;
    y1 += D;
    y2 += D;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2) {
      v[x1].push_back(make_pair(y1, y2));
    } else {
      h[y1].push_back(make_pair(x1, x2));
    }
  }
  for (int y = 0; y < maxn; y++)
    for (int i = 0; i < h[y].size(); i++) {
      pair<int, int> t = h[y][i];
      for (int x = 0; x < maxn; x++) d[x].clear();
      for (int x = t.first; x <= t.second; x++)
        for (int j = 0; j < v[x].size(); j++) {
          pair<int, int> t2 = v[x][j];
          if (t2.first <= y && t2.second > y) {
            add(x, 1);
            d[t2.second].push_back(x);
          }
        }
      for (int y2 = y + 1; y2 < maxn; y2++) {
        for (int j = 0; j < h[y2].size(); j++) {
          pair<int, int> t2 = h[y2][j];
          long long cur = get(t2.first, t2.second);
          ans += cur * (cur - 1) / 2;
        }
        for (int j = 0; j < d[y2].size(); j++) {
          add(d[y2][j], -1);
        }
      }
    }
  printf("%lld\n", ans);
  return 0;
}

```