#include <bits/stdc++.h>
using namespace std;
const int N = 5050;
const int M = 2 * N;
int xl[N], xr[N], yl[N], yr[N];
int sum[M];
void Set(int i, int f) {
  for (; i < M; i += i & -i) sum[i] += f;
}
int Get(int i) {
  int ans = 0;
  for (; i; i -= i & -i) ans += sum[i];
  return ans;
}
vector<int> Us[M];
int main() {
  int n;
  scanf("%i", &n);
  vector<int> hor, ver;
  for (int i = 1; i <= n; i++) {
    scanf("%i %i %i %i", &xl[i], &yl[i], &xr[i], &yr[i]);
    if (xl[i] > xr[i]) swap(xl[i], xr[i]);
    if (yl[i] > yr[i]) swap(yl[i], yr[i]);
    xl[i] += 5001;
    xr[i] += 5001;
    yl[i] += 5001;
    yr[i] += 5001;
    if (xl[i] == xr[i])
      ver.push_back(i);
    else
      hor.push_back(i), Us[xr[i]].push_back(i);
  }
  sort(ver.begin(), ver.end(), [&](int i, int j) { return xl[i] < xl[j]; });
  sort(hor.begin(), hor.end(), [&](int i, int j) { return xr[i] < xr[j]; });
  long long ans = 0;
  for (int i = 0; i + 1 < ver.size(); i++) {
    for (int j = 1; j < M; j++) sum[j] = 0;
    for (int j : hor)
      if (xl[j] <= xl[ver[i]] && xr[j] >= xl[ver[i + 1]]) Set(yl[j], 1);
    for (int j = i + 1, k = 0; j < ver.size(); j++) {
      while (k < hor.size() && xr[hor[k]] < xl[ver[j]]) {
        if (xl[hor[k]] <= xl[ver[i]] && xr[hor[k]] >= xl[ver[i + 1]])
          Set(yl[hor[k]], -1);
        k++;
      }
      int L = max(yl[ver[i]], yl[ver[j]]);
      int R = min(yr[ver[i]], yr[ver[j]]);
      if (L <= R) {
        int cnt = Get(R) - Get(L - 1);
        ans += (long long)cnt * (cnt - 1) / 2;
      }
    }
  }
  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>
using namespace std;
const int N = 5050;
const int M = 2 * N;
int xl[N], xr[N], yl[N], yr[N];
int sum[M];
void Set(int i, int f) {
  for (; i < M; i += i & -i) sum[i] += f;
}
int Get(int i) {
  int ans = 0;
  for (; i; i -= i & -i) ans += sum[i];
  return ans;
}
vector<int> Us[M];
int main() {
  int n;
  scanf("%i", &n);
  vector<int> hor, ver;
  for (int i = 1; i <= n; i++) {
    scanf("%i %i %i %i", &xl[i], &yl[i], &xr[i], &yr[i]);
    if (xl[i] > xr[i]) swap(xl[i], xr[i]);
    if (yl[i] > yr[i]) swap(yl[i], yr[i]);
    xl[i] += 5001;
    xr[i] += 5001;
    yl[i] += 5001;
    yr[i] += 5001;
    if (xl[i] == xr[i])
      ver.push_back(i);
    else
      hor.push_back(i), Us[xr[i]].push_back(i);
  }
  sort(ver.begin(), ver.end(), [&](int i, int j) { return xl[i] < xl[j]; });
  sort(hor.begin(), hor.end(), [&](int i, int j) { return xr[i] < xr[j]; });
  long long ans = 0;
  for (int i = 0; i + 1 < ver.size(); i++) {
    for (int j = 1; j < M; j++) sum[j] = 0;
    for (int j : hor)
      if (xl[j] <= xl[ver[i]] && xr[j] >= xl[ver[i + 1]]) Set(yl[j], 1);
    for (int j = i + 1, k = 0; j < ver.size(); j++) {
      while (k < hor.size() && xr[hor[k]] < xl[ver[j]]) {
        if (xl[hor[k]] <= xl[ver[i]] && xr[hor[k]] >= xl[ver[i + 1]])
          Set(yl[hor[k]], -1);
        k++;
      }
      int L = max(yl[ver[i]], yl[ver[j]]);
      int R = min(yr[ver[i]], yr[ver[j]]);
      if (L <= R) {
        int cnt = Get(R) - Get(L - 1);
        ans += (long long)cnt * (cnt - 1) / 2;
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```