#include <bits/stdc++.h>
#pragma comment(linker, "/STACK: 1024000000,1024000000")
using namespace std;
const int N = 1e4 + 15;
vector<pair<int, int> > h[N], v[N];
vector<int> cur[N];
int c[N];
void add(int x, int p) {
  for (; x < N; x += x & -x) c[x] += p;
}
int qry(int x) {
  int res = 0;
  for (; x; x -= x & -x) res += c[x];
  return res;
}
int sm(int l, int r) { return qry(r) - qry(l - 1); }
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0, x1, y1, x2, y2; i < n; i++) {
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    x1 += 5001;
    x2 += 5001;
    y1 += 5001;
    y2 += 5001;
    if (y1 == y2)
      h[y1].push_back({min(x1, x2), max(x1, x2)});
    else
      v[x1].push_back({min(y1, y2), max(y1, y2)});
  }
  long long ans = 0;
  for (int i = 0; i < N; i++) {
    for (auto nw : h[i]) {
      memset(c, 0, sizeof(c));
      for (int j = 0; j < N; j++) cur[j].clear();
      for (int j = nw.first; j <= nw.second; j++) {
        for (auto nww : v[j]) {
          if (nww.first <= i && nww.second > i) {
            add(j, 1);
            cur[nww.second].push_back(j);
          }
        }
      }
      for (int j = i + 1; j < N; j++) {
        for (auto nww : h[j]) {
          int plus = sm(nww.first, nww.second);
          ans += 1ll * plus * (plus - 1) / 2;
        }
        for (auto nww : cur[j]) add(nww, -1);
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 comment(linker, "/STACK: 1024000000,1024000000")
using namespace std;
const int N = 1e4 + 15;
vector<pair<int, int> > h[N], v[N];
vector<int> cur[N];
int c[N];
void add(int x, int p) {
  for (; x < N; x += x & -x) c[x] += p;
}
int qry(int x) {
  int res = 0;
  for (; x; x -= x & -x) res += c[x];
  return res;
}
int sm(int l, int r) { return qry(r) - qry(l - 1); }
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0, x1, y1, x2, y2; i < n; i++) {
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    x1 += 5001;
    x2 += 5001;
    y1 += 5001;
    y2 += 5001;
    if (y1 == y2)
      h[y1].push_back({min(x1, x2), max(x1, x2)});
    else
      v[x1].push_back({min(y1, y2), max(y1, y2)});
  }
  long long ans = 0;
  for (int i = 0; i < N; i++) {
    for (auto nw : h[i]) {
      memset(c, 0, sizeof(c));
      for (int j = 0; j < N; j++) cur[j].clear();
      for (int j = nw.first; j <= nw.second; j++) {
        for (auto nww : v[j]) {
          if (nww.first <= i && nww.second > i) {
            add(j, 1);
            cur[nww.second].push_back(j);
          }
        }
      }
      for (int j = i + 1; j < N; j++) {
        for (auto nww : h[j]) {
          int plus = sm(nww.first, nww.second);
          ans += 1ll * plus * (plus - 1) / 2;
        }
        for (auto nww : cur[j]) add(nww, -1);
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```