#include <bits/stdc++.h>
using namespace std;
const int mxN = 5e3, mxC = mxN * 2;
int n, ft[mxC + 2];
vector<array<int, 3>> h, v;
void upd(int i, int x) {
  for (++i; i <= mxC + 1; i += i & -i) ft[i] += x;
}
int qry(int i) {
  int r = 0;
  for (; i; i -= i & -i) r += ft[i];
  return r;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0, x1, y1, x2, y2; i < n; ++i) {
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += mxN;
    y1 += mxN;
    x2 += mxN;
    y2 += mxN;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2)
      v.push_back({x1, y1, y2});
    else
      h.push_back({y1, x1, x2});
  }
  sort(h.begin(), h.end());
  sort(v.begin(), v.end(), [](const array<int, 3> &a, const array<int, 3> &b) {
    return a[2] < b[2];
  });
  long long ans = 0;
  for (int i = 0; i < h.size(); ++i) {
    for (int j = 0; j < v.size(); ++j)
      if (v[j][1] <= h[i][0]) upd(v[j][0], 1);
    for (int j1 = i + 1, j2 = 0; j1 < h.size() || j2 < v.size();) {
      if (j2 < v.size() && (j1 >= h.size() || v[j2][2] < h[j1][0])) {
        if (v[j2][1] <= h[i][0]) upd(v[j2][0], -1);
        ++j2;
      } else {
        int l = max(h[i][1], h[j1][1]), r = min(h[i][2], h[j1][2]);
        long long k = l <= r ? qry(r + 1) - qry(l) : 0;
        ans += k * (k - 1) / 2;
        ++j1;
      }
    }
  }
  cout << 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 mxN = 5e3, mxC = mxN * 2;
int n, ft[mxC + 2];
vector<array<int, 3>> h, v;
void upd(int i, int x) {
  for (++i; i <= mxC + 1; i += i & -i) ft[i] += x;
}
int qry(int i) {
  int r = 0;
  for (; i; i -= i & -i) r += ft[i];
  return r;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0, x1, y1, x2, y2; i < n; ++i) {
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += mxN;
    y1 += mxN;
    x2 += mxN;
    y2 += mxN;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2)
      v.push_back({x1, y1, y2});
    else
      h.push_back({y1, x1, x2});
  }
  sort(h.begin(), h.end());
  sort(v.begin(), v.end(), [](const array<int, 3> &a, const array<int, 3> &b) {
    return a[2] < b[2];
  });
  long long ans = 0;
  for (int i = 0; i < h.size(); ++i) {
    for (int j = 0; j < v.size(); ++j)
      if (v[j][1] <= h[i][0]) upd(v[j][0], 1);
    for (int j1 = i + 1, j2 = 0; j1 < h.size() || j2 < v.size();) {
      if (j2 < v.size() && (j1 >= h.size() || v[j2][2] < h[j1][0])) {
        if (v[j2][1] <= h[i][0]) upd(v[j2][0], -1);
        ++j2;
      } else {
        int l = max(h[i][1], h[j1][1]), r = min(h[i][2], h[j1][2]);
        long long k = l <= r ? qry(r + 1) - qry(l) : 0;
        ans += k * (k - 1) / 2;
        ++j1;
      }
    }
  }
  cout << ans;
}

```