#include <bits/stdc++.h>
using namespace std;
const int maxn = 5005;
const int maxl = 1e4 + 2;
int n;
vector<pair<int, int> > hor[maxl], ver[maxl];
vector<int> cur[maxl];
struct node {
  int sum[maxl];
  int lowbit(int x) { return x & (-x); }
  void add(int id, int val) {
    for (int i = id; i < maxl; i += lowbit(i)) {
      sum[i] += val;
    }
  }
  int query(int id) {
    int ans = 0;
    for (int i = id; i >= 1; i -= lowbit(i)) {
      ans += sum[i];
    }
    return ans;
  }
  int s(int l, int r) { return query(r) - query(l - 1); }
  void clear() { memset(sum, 0, sizeof(sum)); }
} bit;
int main() {
  scanf("%d", &n);
  for (int i = 1, x1, x2, y1, y2; i <= n; i++) {
    scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
    x1 += 5001;
    x2 += 5001;
    y1 += 5001;
    y2 += 5001;
    if (y1 == y2)
      hor[y1].push_back(make_pair(min(x1, x2), max(x1, x2)));
    else
      ver[x1].push_back(make_pair(min(y1, y2), max(y1, y2)));
  }
  long long ans = 0;
  for (int i = 1; i < maxl; i++) {
    for (auto h1 : hor[i]) {
      bit.clear();
      for (int j = 1; j < maxl; j++) cur[j].clear();
      for (int j = h1.first; j <= h1.second; j++) {
        for (auto v : ver[j]) {
          if (v.first <= i && v.second > i) {
            cur[v.second].push_back(j);
            bit.add(j, 1);
          }
        }
      }
      for (int y2 = i + 1; y2 < maxl; y2++) {
        for (auto h2 : hor[y2]) {
          int res = bit.s(h2.first, h2.second);
          ans += 1LL * res * (res - 1) / 2;
        }
        for (auto j : cur[y2]) bit.add(j, -1);
      }
    }
  }
  printf("%lld\n", ans);
}

### Prompt

Generate 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 maxn = 5005;
const int maxl = 1e4 + 2;
int n;
vector<pair<int, int> > hor[maxl], ver[maxl];
vector<int> cur[maxl];
struct node {
  int sum[maxl];
  int lowbit(int x) { return x & (-x); }
  void add(int id, int val) {
    for (int i = id; i < maxl; i += lowbit(i)) {
      sum[i] += val;
    }
  }
  int query(int id) {
    int ans = 0;
    for (int i = id; i >= 1; i -= lowbit(i)) {
      ans += sum[i];
    }
    return ans;
  }
  int s(int l, int r) { return query(r) - query(l - 1); }
  void clear() { memset(sum, 0, sizeof(sum)); }
} bit;
int main() {
  scanf("%d", &n);
  for (int i = 1, x1, x2, y1, y2; i <= n; i++) {
    scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
    x1 += 5001;
    x2 += 5001;
    y1 += 5001;
    y2 += 5001;
    if (y1 == y2)
      hor[y1].push_back(make_pair(min(x1, x2), max(x1, x2)));
    else
      ver[x1].push_back(make_pair(min(y1, y2), max(y1, y2)));
  }
  long long ans = 0;
  for (int i = 1; i < maxl; i++) {
    for (auto h1 : hor[i]) {
      bit.clear();
      for (int j = 1; j < maxl; j++) cur[j].clear();
      for (int j = h1.first; j <= h1.second; j++) {
        for (auto v : ver[j]) {
          if (v.first <= i && v.second > i) {
            cur[v.second].push_back(j);
            bit.add(j, 1);
          }
        }
      }
      for (int y2 = i + 1; y2 < maxl; y2++) {
        for (auto h2 : hor[y2]) {
          int res = bit.s(h2.first, h2.second);
          ans += 1LL * res * (res - 1) / 2;
        }
        for (auto j : cur[y2]) bit.add(j, -1);
      }
    }
  }
  printf("%lld\n", ans);
}

```