#include <bits/stdc++.h>
using namespace std;
const int M = 5e3 + 1, N = 1e4 + 1;
vector<array<int, 3> > V, H;
int T[N + 10];
int lowbit(int x) { return x & (-x); }
void add(int x, int val) {
  for (; x <= N; x += lowbit(x)) T[x] += val;
}
int ask(int x) {
  int ans = 0;
  for (; x; x -= lowbit(x)) ans += T[x];
  return ans;
}
int main() {
  int n, x1, y1, x2, y2;
  cin >> n;
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    x1 += M;
    x2 += M;
    y1 += M;
    y2 += M;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2) V.push_back({x1, y1, y2});
    if (y1 == y2) H.push_back({y1, x1, x2});
  }
  sort(H.begin(), H.end());
  sort(V.begin(), V.end(),
       [](array<int, 3> a, array<int, 3> b) { return a[2] < b[2]; });
  for (int i = 0; i < H.size(); ++i) {
    memset(T, 0, sizeof T);
    for (int j = 0; j < V.size(); ++j)
      if (V[j][1] <= H[i][0]) add(V[j][0], 1);
    for (int j = i + 1, k = 0; j < H.size();) {
      if (k < V.size() && V[k][2] < H[j][0]) {
        if (V[k][1] <= H[i][0]) add(V[k][0], -1);
        ++k;
      } else {
        int l = max(H[i][1], H[j][1]);
        int r = min(H[i][2], H[j][2]);
        if (l < r) {
          int sum = ask(r) - ask(l - 1);
          ans = ans + sum * (sum - 1) / 2;
        }
        ++j;
      }
    }
  }
  cout << ans << endl;
}

### Prompt

Create a solution in CPP for 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 M = 5e3 + 1, N = 1e4 + 1;
vector<array<int, 3> > V, H;
int T[N + 10];
int lowbit(int x) { return x & (-x); }
void add(int x, int val) {
  for (; x <= N; x += lowbit(x)) T[x] += val;
}
int ask(int x) {
  int ans = 0;
  for (; x; x -= lowbit(x)) ans += T[x];
  return ans;
}
int main() {
  int n, x1, y1, x2, y2;
  cin >> n;
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    x1 += M;
    x2 += M;
    y1 += M;
    y2 += M;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (x1 == x2) V.push_back({x1, y1, y2});
    if (y1 == y2) H.push_back({y1, x1, x2});
  }
  sort(H.begin(), H.end());
  sort(V.begin(), V.end(),
       [](array<int, 3> a, array<int, 3> b) { return a[2] < b[2]; });
  for (int i = 0; i < H.size(); ++i) {
    memset(T, 0, sizeof T);
    for (int j = 0; j < V.size(); ++j)
      if (V[j][1] <= H[i][0]) add(V[j][0], 1);
    for (int j = i + 1, k = 0; j < H.size();) {
      if (k < V.size() && V[k][2] < H[j][0]) {
        if (V[k][1] <= H[i][0]) add(V[k][0], -1);
        ++k;
      } else {
        int l = max(H[i][1], H[j][1]);
        int r = min(H[i][2], H[j][2]);
        if (l < r) {
          int sum = ask(r) - ask(l - 1);
          ans = ans + sum * (sum - 1) / 2;
        }
        ++j;
      }
    }
  }
  cout << ans << endl;
}

```