#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e4 + 10;
int n, m, gs1 = 0, gs2 = 0, c[maxn], vis[maxn];
long long ans = 0;
struct Node1 {
  int x, y, xx;
  bool operator<(const Node1 &rhs) const { return y < rhs.y; }
} a[maxn];
struct Node2 {
  int x, y, yy;
  bool operator<(const Node2 &rhs) const { return yy < rhs.yy; }
} b[maxn];
long long lowbit(long long x) { return x & -x; }
long long find(long long x) {
  long long ret = 0;
  while (x > 0) {
    ret += c[x];
    x -= lowbit(x);
  }
  return ret;
}
void add(int x, int val) {
  while (x <= 10001) {
    c[x] += val;
    x += lowbit(x);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int x, y, xx, yy;
    scanf("%d%d%d%d", &x, &y, &xx, &yy);
    x += 5001, y += 5001, xx += 5001, yy += 5001;
    if (x == xx)
      b[++gs2] = (Node2){x, min(y, yy), max(y, yy)};
    else
      a[++gs1] = (Node1){min(x, xx), y, max(x, xx)};
  }
  sort(a + 1, a + 1 + gs1);
  sort(b + 1, b + 1 + gs2);
  sort(a + 1, a + 1 + gs1);
  sort(b + 1, b + 1 + gs2);
  for (int i = 1; i <= gs1; i++) {
    memset(c, 0, sizeof(c));
    for (int j = 1; j <= gs2; j++)
      if (b[j].y <= a[i].y && b[j].yy >= a[i].y) {
        add(b[j].x, 1);
        vis[j] = 1;
      } else
        vis[j] = 0;
    int k = 1;
    for (int j = i + 1; j <= gs1; j++) {
      while (k <= gs2 && b[k].yy < a[j].y) {
        if (vis[k]) add(b[k].x, -1);
        k++;
      }
      int r = min(a[i].xx, a[j].xx), l = max(a[i].x, a[j].x) - 1;
      if (l < r) {
        int x = find(r) - find(l);
        ans += 1ll * x * (x - 1) / 2;
      }
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Your task is to create 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;
const int maxn = 1e4 + 10;
int n, m, gs1 = 0, gs2 = 0, c[maxn], vis[maxn];
long long ans = 0;
struct Node1 {
  int x, y, xx;
  bool operator<(const Node1 &rhs) const { return y < rhs.y; }
} a[maxn];
struct Node2 {
  int x, y, yy;
  bool operator<(const Node2 &rhs) const { return yy < rhs.yy; }
} b[maxn];
long long lowbit(long long x) { return x & -x; }
long long find(long long x) {
  long long ret = 0;
  while (x > 0) {
    ret += c[x];
    x -= lowbit(x);
  }
  return ret;
}
void add(int x, int val) {
  while (x <= 10001) {
    c[x] += val;
    x += lowbit(x);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int x, y, xx, yy;
    scanf("%d%d%d%d", &x, &y, &xx, &yy);
    x += 5001, y += 5001, xx += 5001, yy += 5001;
    if (x == xx)
      b[++gs2] = (Node2){x, min(y, yy), max(y, yy)};
    else
      a[++gs1] = (Node1){min(x, xx), y, max(x, xx)};
  }
  sort(a + 1, a + 1 + gs1);
  sort(b + 1, b + 1 + gs2);
  sort(a + 1, a + 1 + gs1);
  sort(b + 1, b + 1 + gs2);
  for (int i = 1; i <= gs1; i++) {
    memset(c, 0, sizeof(c));
    for (int j = 1; j <= gs2; j++)
      if (b[j].y <= a[i].y && b[j].yy >= a[i].y) {
        add(b[j].x, 1);
        vis[j] = 1;
      } else
        vis[j] = 0;
    int k = 1;
    for (int j = i + 1; j <= gs1; j++) {
      while (k <= gs2 && b[k].yy < a[j].y) {
        if (vis[k]) add(b[k].x, -1);
        k++;
      }
      int r = min(a[i].xx, a[j].xx), l = max(a[i].x, a[j].x) - 1;
      if (l < r) {
        int x = find(r) - find(l);
        ans += 1ll * x * (x - 1) / 2;
      }
    }
  }
  cout << ans;
  return 0;
}

```