#include <bits/stdc++.h>
using namespace std;
const int mx = 5e3 + 10;
const int MX = 1e4 + 10;
int n, m;
struct Node {
  int x, y, x2, y2;
} eg[mx], eg2[mx];
bool cmp(const Node &a, const Node &b) { return a.y2 < b.y2; }
int cnt, cnt2;
int s[mx], tot;
struct Tree {
  int a[MX], up = 1e4 + 1;
  int lowbit(int x) { return x & -x; }
  void updata(int x, int c) {
    while (x <= up) {
      a[x] += c;
      x += lowbit(x);
    }
  }
  int query(int x) {
    int ans = 0;
    while (x) {
      ans += a[x];
      x -= lowbit(x);
    }
    return ans;
  }
} t;
int main() {
  long long ans = 0;
  int x, y, x2, y2;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d%d%d", &x, &y, &x2, &y2);
    x += 5001, y += 5001, x2 += 5001, y2 += 5001;
    if (y == y2) {
      if (x > x2) swap(x, x2);
      eg[++cnt] = Node{x, y, x2, y2};
    } else {
      if (y > y2) swap(y, y2);
      eg2[++cnt2] = Node{x, y, x2, y2};
    }
  }
  sort(eg + 1, eg + cnt + 1, cmp);
  sort(eg2 + 1, eg2 + cnt2 + 1, cmp);
  for (int i = 1; i <= cnt; ++i) {
    for (int j = cnt2; j >= 1; --j) {
      if (eg2[j].y <= eg[i].y && eg2[j].y2 >= eg[i].y)
        s[++tot] = j, t.updata(eg2[j].x, 1);
    }
    for (int j = i + 1; j <= cnt; ++j) {
      while (tot && eg2[s[tot]].y2 < eg[j].y)
        t.updata(eg2[s[tot]].x, -1), --tot;
      int l = max(eg[i].x, eg[j].x);
      int r = min(eg[i].x2, eg[j].x2);
      if (l <= r) {
        int ah = t.query(r) - t.query(l - 1);
        ans += ah * (ah - 1) / 2;
      }
    }
    while (tot) t.updata(eg2[s[tot--]].x, -1);
  }
  printf("%I64d\n", ans);
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 mx = 5e3 + 10;
const int MX = 1e4 + 10;
int n, m;
struct Node {
  int x, y, x2, y2;
} eg[mx], eg2[mx];
bool cmp(const Node &a, const Node &b) { return a.y2 < b.y2; }
int cnt, cnt2;
int s[mx], tot;
struct Tree {
  int a[MX], up = 1e4 + 1;
  int lowbit(int x) { return x & -x; }
  void updata(int x, int c) {
    while (x <= up) {
      a[x] += c;
      x += lowbit(x);
    }
  }
  int query(int x) {
    int ans = 0;
    while (x) {
      ans += a[x];
      x -= lowbit(x);
    }
    return ans;
  }
} t;
int main() {
  long long ans = 0;
  int x, y, x2, y2;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d%d%d%d", &x, &y, &x2, &y2);
    x += 5001, y += 5001, x2 += 5001, y2 += 5001;
    if (y == y2) {
      if (x > x2) swap(x, x2);
      eg[++cnt] = Node{x, y, x2, y2};
    } else {
      if (y > y2) swap(y, y2);
      eg2[++cnt2] = Node{x, y, x2, y2};
    }
  }
  sort(eg + 1, eg + cnt + 1, cmp);
  sort(eg2 + 1, eg2 + cnt2 + 1, cmp);
  for (int i = 1; i <= cnt; ++i) {
    for (int j = cnt2; j >= 1; --j) {
      if (eg2[j].y <= eg[i].y && eg2[j].y2 >= eg[i].y)
        s[++tot] = j, t.updata(eg2[j].x, 1);
    }
    for (int j = i + 1; j <= cnt; ++j) {
      while (tot && eg2[s[tot]].y2 < eg[j].y)
        t.updata(eg2[s[tot]].x, -1), --tot;
      int l = max(eg[i].x, eg[j].x);
      int r = min(eg[i].x2, eg[j].x2);
      if (l <= r) {
        int ah = t.query(r) - t.query(l - 1);
        ans += ah * (ah - 1) / 2;
      }
    }
    while (tot) t.updata(eg2[s[tot--]].x, -1);
  }
  printf("%I64d\n", ans);
}

```