#include <bits/stdc++.h>
using namespace std;
struct aaa {
  int l1, r1, l2, r2;
} a[10001], b[10001], c[10001];
int n, cnt1, cnt2, v[10010], f[10010];
long long ans;
bool cmp(aaa a, aaa b) { return a.r1 < b.r1; }
bool cmp1(aaa a, aaa b) { return a.r2 < b.r2; }
int lowbit(int x) { return x & (-x); }
void ins(int x, int t) {
  int i;
  for (i = x; i <= 10001; i += lowbit(i)) f[i] += t;
}
int query(int x) {
  int xlh = 0, i;
  for (i = x; i; i -= lowbit(i)) xlh += f[i];
  return xlh;
}
int main() {
  int i, j, l, r, la;
  long long now;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d%d%d%d", &a[i].l1, &a[i].r1, &a[i].l2, &a[i].r2);
    if (a[i].r1 == a[i].r2) {
      b[++cnt1] = a[i];
      if (b[cnt1].l1 > b[cnt1].l2) swap(b[cnt1].l1, b[cnt1].l2);
    } else {
      c[++cnt2] = a[i];
      if (c[cnt2].r1 > c[cnt2].r2) swap(c[cnt2].r1, c[cnt2].r2);
    }
  }
  sort(b + 1, b + cnt1 + 1, cmp);
  sort(c + 1, c + cnt2 + 1, cmp1);
  for (i = 1; i <= cnt1; i++) {
    la = 1;
    memset(f, 0, sizeof(f));
    for (j = 1; j <= cnt2; j++)
      if (c[j].r1 <= b[i].r1 && c[j].r2 >= b[i].r1)
        ins(c[j].l1 + 5001, 1), v[j] = 1;
      else
        v[j] = 0;
    for (j = i + 1; j <= cnt1; j++) {
      while (la <= cnt2 && c[la].r2 < b[j].r1) {
        if (v[la]) ins(c[la].l1 + 5001, -1);
        la++;
      }
      l = max(b[i].l1, b[j].l1);
      r = min(b[i].l2, b[j].l2);
      if (l <= r) {
        now = query(r + 5001) - query(l - 1 + 5001);
        ans += now * (now - 1) / 2ll;
      }
    }
  }
  printf("%lld", ans);
}

### Prompt

Please formulate 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;
struct aaa {
  int l1, r1, l2, r2;
} a[10001], b[10001], c[10001];
int n, cnt1, cnt2, v[10010], f[10010];
long long ans;
bool cmp(aaa a, aaa b) { return a.r1 < b.r1; }
bool cmp1(aaa a, aaa b) { return a.r2 < b.r2; }
int lowbit(int x) { return x & (-x); }
void ins(int x, int t) {
  int i;
  for (i = x; i <= 10001; i += lowbit(i)) f[i] += t;
}
int query(int x) {
  int xlh = 0, i;
  for (i = x; i; i -= lowbit(i)) xlh += f[i];
  return xlh;
}
int main() {
  int i, j, l, r, la;
  long long now;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d%d%d%d", &a[i].l1, &a[i].r1, &a[i].l2, &a[i].r2);
    if (a[i].r1 == a[i].r2) {
      b[++cnt1] = a[i];
      if (b[cnt1].l1 > b[cnt1].l2) swap(b[cnt1].l1, b[cnt1].l2);
    } else {
      c[++cnt2] = a[i];
      if (c[cnt2].r1 > c[cnt2].r2) swap(c[cnt2].r1, c[cnt2].r2);
    }
  }
  sort(b + 1, b + cnt1 + 1, cmp);
  sort(c + 1, c + cnt2 + 1, cmp1);
  for (i = 1; i <= cnt1; i++) {
    la = 1;
    memset(f, 0, sizeof(f));
    for (j = 1; j <= cnt2; j++)
      if (c[j].r1 <= b[i].r1 && c[j].r2 >= b[i].r1)
        ins(c[j].l1 + 5001, 1), v[j] = 1;
      else
        v[j] = 0;
    for (j = i + 1; j <= cnt1; j++) {
      while (la <= cnt2 && c[la].r2 < b[j].r1) {
        if (v[la]) ins(c[la].l1 + 5001, -1);
        la++;
      }
      l = max(b[i].l1, b[j].l1);
      r = min(b[i].l2, b[j].l2);
      if (l <= r) {
        now = query(r + 5001) - query(l - 1 + 5001);
        ans += now * (now - 1) / 2ll;
      }
    }
  }
  printf("%lld", ans);
}

```