#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e3 + 5;
const int N = 1e4 + 5;
const int inf = 0x3f3f3f3f;
struct hor {
  int x1, x2, y;
  bool operator<(const hor& a) const { return y < a.y; }
} h[maxn];
struct ver {
  int y1, y2, x;
  bool operator<(const ver& a) const { return y1 < a.y1; }
} v[maxn];
int n, cntv, cnth;
int c[N];
int lowbit(int x) { return x & (-x); }
int sum(int x) {
  int ans = 0;
  for (int i = x; i > 0; i -= lowbit(i)) ans += c[i];
  return ans;
}
int get_sum(int l, int r) { return sum(r) - sum(l - 1); }
void update(int x, int val) {
  for (int i = x; i <= N; i += lowbit(i)) c[i] += val;
}
int main() {
  scanf("%d", &n);
  cntv = cnth = 0;
  for (int i = 1; i <= n; ++i) {
    int x1, y1, x2, y2;
    scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
    x1 += 5001, y1 += 5001, x2 += 5001, y2 += 5001;
    if (x1 == x2)
      v[++cntv] = {min(y1, y2), max(y1, y2), x1};
    else
      h[++cnth] = {min(x1, x2), max(x1, x2), y1};
  }
  sort(h + 1, h + cnth + 1);
  sort(v + 1, v + cntv + 1);
  long long ans = 0;
  for (int i = 1; i <= cnth; ++i) {
    memset(c, 0, sizeof(c));
    for (int j = 1, k = 1; k < i;) {
      if (v[j].y1 <= h[k].y && j <= cntv) {
        if (v[j].x >= h[i].x1 && v[j].x <= h[i].x2 && v[j].y2 >= h[i].y)
          update(v[j].x, 1);
        ++j;
      } else {
        int res = get_sum(h[k].x1, h[k].x2);
        ans += 1ll * res * (res - 1) / 2;
        ++k;
      }
    }
  }
  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;
const int maxn = 5e3 + 5;
const int N = 1e4 + 5;
const int inf = 0x3f3f3f3f;
struct hor {
  int x1, x2, y;
  bool operator<(const hor& a) const { return y < a.y; }
} h[maxn];
struct ver {
  int y1, y2, x;
  bool operator<(const ver& a) const { return y1 < a.y1; }
} v[maxn];
int n, cntv, cnth;
int c[N];
int lowbit(int x) { return x & (-x); }
int sum(int x) {
  int ans = 0;
  for (int i = x; i > 0; i -= lowbit(i)) ans += c[i];
  return ans;
}
int get_sum(int l, int r) { return sum(r) - sum(l - 1); }
void update(int x, int val) {
  for (int i = x; i <= N; i += lowbit(i)) c[i] += val;
}
int main() {
  scanf("%d", &n);
  cntv = cnth = 0;
  for (int i = 1; i <= n; ++i) {
    int x1, y1, x2, y2;
    scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
    x1 += 5001, y1 += 5001, x2 += 5001, y2 += 5001;
    if (x1 == x2)
      v[++cntv] = {min(y1, y2), max(y1, y2), x1};
    else
      h[++cnth] = {min(x1, x2), max(x1, x2), y1};
  }
  sort(h + 1, h + cnth + 1);
  sort(v + 1, v + cntv + 1);
  long long ans = 0;
  for (int i = 1; i <= cnth; ++i) {
    memset(c, 0, sizeof(c));
    for (int j = 1, k = 1; k < i;) {
      if (v[j].y1 <= h[k].y && j <= cntv) {
        if (v[j].x >= h[i].x1 && v[j].x <= h[i].x2 && v[j].y2 >= h[i].y)
          update(v[j].x, 1);
        ++j;
      } else {
        int res = get_sum(h[k].x1, h[k].x2);
        ans += 1ll * res * (res - 1) / 2;
        ++k;
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```