#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e4 + 10;
struct node {
  int x, y, xx, yy;
} a[maxn], b[maxn], e[maxn];
int tree[maxn];
int bb[maxn * 4];
map<int, int> mp;
bool cmp1(node p, node q) { return p.y < q.y; }
bool cmp2(node p, node q) { return p.yy < q.yy; }
int lowbit(int x) { return x & -x; }
void add(int x, int val) {
  while (x < maxn) {
    tree[x] += val;
    x += lowbit(x);
  }
}
int query(int x) {
  int res = 0;
  while (x > 0) {
    res += tree[x];
    x -= lowbit(x);
  }
  return res;
}
int main() {
  int n, x, y, xx, yy;
  scanf("%d", &n);
  int cnt = 0, num1 = 0, num2 = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d%d%d%d", &x, &y, &xx, &yy);
    if (y == yy && x > xx) swap(x, xx);
    if (x == xx && y > yy) swap(y, yy);
    bb[++cnt] = x;
    bb[++cnt] = xx, bb[++cnt] = y;
    bb[++cnt] = yy;
    if (y == yy)
      a[++num1] = {x, y, xx, yy};
    else
      b[++num2] = {x, y, xx, yy};
  }
  sort(bb + 1, bb + 1 + cnt);
  int m = unique(bb + 1, bb + 1 + cnt) - bb - 1;
  for (int i = 1; i <= m; i++) mp[bb[i]] = i;
  for (int i = 1; i <= num1; i++) {
    a[i].x = mp[a[i].x];
    a[i].xx = mp[a[i].xx];
    a[i].y = mp[a[i].y];
    a[i].yy = mp[a[i].yy];
  }
  for (int i = 1; i <= num2; i++) {
    b[i].x = mp[b[i].x];
    b[i].xx = mp[b[i].xx];
    b[i].y = mp[b[i].y];
    b[i].yy = mp[b[i].yy];
  }
  sort(a + 1, a + 1 + num1, cmp1);
  sort(b + 1, b + 1 + num2, cmp2);
  long long ans = 0;
  for (int i = 1; i <= num1; i++) {
    int num = 0, tmp = 1;
    for (int j = 1; j <= num2; j++)
      if (b[j].y <= a[i].y) {
        add(b[j].x, 1);
        e[++num] = b[j];
      }
    for (int j = i + 1; j <= num1; j++) {
      while (tmp <= num && e[tmp].yy < a[j].y) {
        add(e[tmp].x, -1);
        tmp++;
      }
      int l = max(a[i].x, a[j].x), r = min(a[i].xx, a[j].xx);
      if (l <= r) {
        int num = query(r) - query(l - 1);
        ans += 1ll * num * (num - 1) / 2;
      }
    }
    while (tmp <= num) add(e[tmp++].x, -1);
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 = 2e4 + 10;
struct node {
  int x, y, xx, yy;
} a[maxn], b[maxn], e[maxn];
int tree[maxn];
int bb[maxn * 4];
map<int, int> mp;
bool cmp1(node p, node q) { return p.y < q.y; }
bool cmp2(node p, node q) { return p.yy < q.yy; }
int lowbit(int x) { return x & -x; }
void add(int x, int val) {
  while (x < maxn) {
    tree[x] += val;
    x += lowbit(x);
  }
}
int query(int x) {
  int res = 0;
  while (x > 0) {
    res += tree[x];
    x -= lowbit(x);
  }
  return res;
}
int main() {
  int n, x, y, xx, yy;
  scanf("%d", &n);
  int cnt = 0, num1 = 0, num2 = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d%d%d%d", &x, &y, &xx, &yy);
    if (y == yy && x > xx) swap(x, xx);
    if (x == xx && y > yy) swap(y, yy);
    bb[++cnt] = x;
    bb[++cnt] = xx, bb[++cnt] = y;
    bb[++cnt] = yy;
    if (y == yy)
      a[++num1] = {x, y, xx, yy};
    else
      b[++num2] = {x, y, xx, yy};
  }
  sort(bb + 1, bb + 1 + cnt);
  int m = unique(bb + 1, bb + 1 + cnt) - bb - 1;
  for (int i = 1; i <= m; i++) mp[bb[i]] = i;
  for (int i = 1; i <= num1; i++) {
    a[i].x = mp[a[i].x];
    a[i].xx = mp[a[i].xx];
    a[i].y = mp[a[i].y];
    a[i].yy = mp[a[i].yy];
  }
  for (int i = 1; i <= num2; i++) {
    b[i].x = mp[b[i].x];
    b[i].xx = mp[b[i].xx];
    b[i].y = mp[b[i].y];
    b[i].yy = mp[b[i].yy];
  }
  sort(a + 1, a + 1 + num1, cmp1);
  sort(b + 1, b + 1 + num2, cmp2);
  long long ans = 0;
  for (int i = 1; i <= num1; i++) {
    int num = 0, tmp = 1;
    for (int j = 1; j <= num2; j++)
      if (b[j].y <= a[i].y) {
        add(b[j].x, 1);
        e[++num] = b[j];
      }
    for (int j = i + 1; j <= num1; j++) {
      while (tmp <= num && e[tmp].yy < a[j].y) {
        add(e[tmp].x, -1);
        tmp++;
      }
      int l = max(a[i].x, a[j].x), r = min(a[i].xx, a[j].xx);
      if (l <= r) {
        int num = query(r) - query(l - 1);
        ans += 1ll * num * (num - 1) / 2;
      }
    }
    while (tmp <= num) add(e[tmp++].x, -1);
  }
  printf("%lld\n", ans);
  return 0;
}

```