#include <bits/stdc++.h>
using namespace std;
int n, p, q, c[5005 << 1], T = (5005 << 1) - 5, bas = 5002;
long long ans = 0;
struct node {
  int x, l, r;
} a[5005], b[5005];
bool cmp(node x, node y) { return x.x < y.x; }
bool cmp_(node x, node y) { return x.l < y.l; }
inline void add(int x) {
  for (; x <= T; x += x & -x) c[x]++;
}
inline int ask(int x) {
  int sum = 0;
  for (; x; x -= x & -x) sum += c[x];
  return sum;
}
void calc(int x) {
  memset(c, 0, sizeof(c));
  int j = 1;
  for (int i = 1; i <= x - 1; i++) {
    while (j <= q && b[j].l <= a[i].x) {
      if (b[j].r >= a[x].x && b[j].x <= a[x].r && b[j].x >= a[x].l)
        add(bas + b[j].x);
      j++;
    }
    int cur = ask(a[i].r + bas) - ask(a[i].l + bas - 1);
    ans += cur * (cur - 1) / 2;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x, y, X, Y;
    scanf("%d%d%d%d", &x, &y, &X, &Y);
    if (x > X) swap(x, X);
    if (y > Y) swap(y, Y);
    if (x == X)
      a[++p].x = x, a[p].l = y, a[p].r = Y;
    else
      b[++q].x = y, b[q].l = x, b[q].r = X;
  }
  sort(a + 1, a + p + 1, cmp);
  sort(b + 1, b + q + 1, cmp_);
  for (int i = 2; i <= p; i++) calc(i);
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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, p, q, c[5005 << 1], T = (5005 << 1) - 5, bas = 5002;
long long ans = 0;
struct node {
  int x, l, r;
} a[5005], b[5005];
bool cmp(node x, node y) { return x.x < y.x; }
bool cmp_(node x, node y) { return x.l < y.l; }
inline void add(int x) {
  for (; x <= T; x += x & -x) c[x]++;
}
inline int ask(int x) {
  int sum = 0;
  for (; x; x -= x & -x) sum += c[x];
  return sum;
}
void calc(int x) {
  memset(c, 0, sizeof(c));
  int j = 1;
  for (int i = 1; i <= x - 1; i++) {
    while (j <= q && b[j].l <= a[i].x) {
      if (b[j].r >= a[x].x && b[j].x <= a[x].r && b[j].x >= a[x].l)
        add(bas + b[j].x);
      j++;
    }
    int cur = ask(a[i].r + bas) - ask(a[i].l + bas - 1);
    ans += cur * (cur - 1) / 2;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x, y, X, Y;
    scanf("%d%d%d%d", &x, &y, &X, &Y);
    if (x > X) swap(x, X);
    if (y > Y) swap(y, Y);
    if (x == X)
      a[++p].x = x, a[p].l = y, a[p].r = Y;
    else
      b[++q].x = y, b[q].l = x, b[q].r = X;
  }
  sort(a + 1, a + p + 1, cmp);
  sort(b + 1, b + q + 1, cmp_);
  for (int i = 2; i <= p; i++) calc(i);
  printf("%lld\n", ans);
  return 0;
}

```