#include <bits/stdc++.h>
using namespace std;
int n, tot, num, c[10001 + 5], q[5000 + 5];
long long ans;
struct node {
  int x1, y1, x2, y2;
  bool operator<(const node &x) const { return y2 < x.y2; }
} R[5000 + 10], C[5000 + 10];
void add(int x, int y) {
  for (; x <= 10001; x += x & -x) c[x] += y;
}
int query(int x, int res = 0) {
  for (; x; x -= x & -x) res += c[x];
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x1, y1, x2, y2;
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    x1 += 5000 + 1, y1 += 5000 + 1, x2 += 5000 + 1, y2 += 5000 + 1;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (y1 == y2)
      R[++tot] = (node){x1, y1, x2, y2};
    else
      C[++num] = (node){x1, y1, x2, y2};
  }
  sort(R + 1, R + tot + 1);
  sort(C + 1, C + num + 1);
  for (int i = 1; i <= tot; i++) {
    int t = 0, k = 1, res, l, r;
    memset(c, 0, sizeof(c));
    for (int j = 1; j <= num; j++)
      if (R[i].x1 <= C[j].x1 && R[i].x2 >= C[j].x1 && C[j].y1 <= R[i].y1 &&
          C[j].y2 >= R[i].y2)
        add(C[j].x1, 1), q[++t] = j;
    for (int j = i + 1; j <= tot; j++) {
      for (; C[q[k]].y2 < R[j].y1 && k <= t; k++) add(C[q[k]].x1, -1);
      l = max(R[i].x1, R[j].x1), r = min(R[i].x2, R[j].x2);
      if (l < r)
        res = query(r) - query(l - 1), ans += 1ll * res * (res - 1) / 2;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### 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;
int n, tot, num, c[10001 + 5], q[5000 + 5];
long long ans;
struct node {
  int x1, y1, x2, y2;
  bool operator<(const node &x) const { return y2 < x.y2; }
} R[5000 + 10], C[5000 + 10];
void add(int x, int y) {
  for (; x <= 10001; x += x & -x) c[x] += y;
}
int query(int x, int res = 0) {
  for (; x; x -= x & -x) res += c[x];
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x1, y1, x2, y2;
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    x1 += 5000 + 1, y1 += 5000 + 1, x2 += 5000 + 1, y2 += 5000 + 1;
    if (x1 > x2) swap(x1, x2);
    if (y1 > y2) swap(y1, y2);
    if (y1 == y2)
      R[++tot] = (node){x1, y1, x2, y2};
    else
      C[++num] = (node){x1, y1, x2, y2};
  }
  sort(R + 1, R + tot + 1);
  sort(C + 1, C + num + 1);
  for (int i = 1; i <= tot; i++) {
    int t = 0, k = 1, res, l, r;
    memset(c, 0, sizeof(c));
    for (int j = 1; j <= num; j++)
      if (R[i].x1 <= C[j].x1 && R[i].x2 >= C[j].x1 && C[j].y1 <= R[i].y1 &&
          C[j].y2 >= R[i].y2)
        add(C[j].x1, 1), q[++t] = j;
    for (int j = i + 1; j <= tot; j++) {
      for (; C[q[k]].y2 < R[j].y1 && k <= t; k++) add(C[q[k]].x1, -1);
      l = max(R[i].x1, R[j].x1), r = min(R[i].x2, R[j].x2);
      if (l < r)
        res = query(r) - query(l - 1), ans += 1ll * res * (res - 1) / 2;
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```