#include <bits/stdc++.h>
using namespace std;
const int maxn = 200000;
long long tr[maxn + 100];
struct node {
  int x, l, r;
};
node q[200000], p[200000];
int n;
void add(int x, long long t) {
  while (x <= maxn) {
    tr[x] += t;
    x += x & (-x);
  }
}
long long query(int x) {
  long long ans = 0;
  while (x > 0) {
    ans += tr[x];
    x -= x & (-x);
  }
  return ans;
}
bool cmp(node a, node b) { return a.x < b.x; }
struct nod {
  int x, r;
} t[maxn];
bool cmp1(nod a, nod b) { return a.r < b.r; }
int main() {
  scanf("%d", &n);
  int x1, x2, y1, y2, cp = 0, cq = 0;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    x1 += 5001;
    x2 += 5001;
    y2 += 5001;
    y1 += 5001;
    if (x1 == x2) p[++cp].x = x1, p[cp].l = min(y1, y2), p[cp].r = max(y1, y2);
    if (y1 == y2) q[++cq].x = y1, q[cq].l = min(x1, x2), q[cq].r = max(x1, x2);
  }
  sort(p + 1, p + 1 + cp, cmp);
  sort(q + 1, q + 1 + cq, cmp);
  for (int i = 1; i <= cp; i++) {
    int top = 0;
    for (int j = 1; j <= cq; j++) {
      if (q[j].l <= p[i].x && q[j].r >= p[i].x && q[j].x >= p[i].l &&
          q[j].x <= p[i].r) {
        t[++top] = (nod){q[j].x, q[j].r}, add(q[j].x, 1);
      }
    }
    sort(t + 1, t + 1 + top, cmp1);
    int nw = 0;
    for (int j = i + 1; j <= cp; j++) {
      if (p[j].r < p[i].l || p[j].l > p[i].r) continue;
      while (nw < top && t[nw + 1].r < p[j].x) {
        nw++;
        add(t[nw].x, -1);
      }
      long long tt = query(p[j].r) - query(p[j].l - 1);
      ans += (long long)tt * (tt - 1) / 2;
    }
    while (nw < top) {
      nw++;
      add(t[nw].x, -1);
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Create a solution in Cpp for 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 = 200000;
long long tr[maxn + 100];
struct node {
  int x, l, r;
};
node q[200000], p[200000];
int n;
void add(int x, long long t) {
  while (x <= maxn) {
    tr[x] += t;
    x += x & (-x);
  }
}
long long query(int x) {
  long long ans = 0;
  while (x > 0) {
    ans += tr[x];
    x -= x & (-x);
  }
  return ans;
}
bool cmp(node a, node b) { return a.x < b.x; }
struct nod {
  int x, r;
} t[maxn];
bool cmp1(nod a, nod b) { return a.r < b.r; }
int main() {
  scanf("%d", &n);
  int x1, x2, y1, y2, cp = 0, cq = 0;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    x1 += 5001;
    x2 += 5001;
    y2 += 5001;
    y1 += 5001;
    if (x1 == x2) p[++cp].x = x1, p[cp].l = min(y1, y2), p[cp].r = max(y1, y2);
    if (y1 == y2) q[++cq].x = y1, q[cq].l = min(x1, x2), q[cq].r = max(x1, x2);
  }
  sort(p + 1, p + 1 + cp, cmp);
  sort(q + 1, q + 1 + cq, cmp);
  for (int i = 1; i <= cp; i++) {
    int top = 0;
    for (int j = 1; j <= cq; j++) {
      if (q[j].l <= p[i].x && q[j].r >= p[i].x && q[j].x >= p[i].l &&
          q[j].x <= p[i].r) {
        t[++top] = (nod){q[j].x, q[j].r}, add(q[j].x, 1);
      }
    }
    sort(t + 1, t + 1 + top, cmp1);
    int nw = 0;
    for (int j = i + 1; j <= cp; j++) {
      if (p[j].r < p[i].l || p[j].l > p[i].r) continue;
      while (nw < top && t[nw + 1].r < p[j].x) {
        nw++;
        add(t[nw].x, -1);
      }
      long long tt = query(p[j].r) - query(p[j].l - 1);
      ans += (long long)tt * (tt - 1) / 2;
    }
    while (nw < top) {
      nw++;
      add(t[nw].x, -1);
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```