#include <bits/stdc++.h>
using namespace std;
const int maxn = 10000 + 233;
int c[maxn];
int n;
bool vis[maxn];
int lowbit(int x) { return x & (-x); }
void add(int x, int d) {
  while (x < maxn) {
    c[x] += d;
    x += lowbit(x);
  }
}
int sum(int x) {
  int ret = 1;
  while (x) {
    ret += c[x];
    x -= lowbit(x);
  }
  return ret;
}
struct node {
  int l, r, p;
} x[maxn], y[maxn];
bool cmpx(node &x, node &y) { return x.p < y.p; }
bool cmpy(node &x, node &y) { return x.r < y.r; }
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  int xcnt = 0, ycnt = 0;
  for (int i = 1; i <= n; i++) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    a += 5001, b += 5001, c += 5001, d += 5001;
    if (a > c) swap(a, c);
    if (b > d) swap(b, d);
    if (a == c) {
      y[++ycnt] = node{b, d, a};
    }
    if (b == d) {
      x[++xcnt] = node{a, c, b};
    }
  }
  sort(x + 1, x + xcnt + 1, cmpx);
  sort(y + 1, y + ycnt + 1, cmpy);
  long long res = 0;
  for (int i = 1; i <= xcnt; i++) {
    memset(vis, 0, sizeof(vis));
    memset(c, 0, sizeof(c));
    for (int j = 1; j <= ycnt; j++) {
      if (y[j].l <= x[i].p && y[j].p >= x[i].l && y[j].p <= x[i].r) {
        vis[j] = 1;
        add(y[j].p, 1);
      }
    }
    int k = 1;
    for (int j = i + 1; j <= xcnt; j++) {
      while (k <= ycnt && y[k].r < x[j].p) {
        if (vis[k]) add(y[k].p, -1);
        k++;
      }
      long long tmp = sum(min(x[i].r, x[j].r)) - sum(max(x[i].l, x[j].l) - 1);
      res += 1ll * tmp * (tmp - 1) / 2;
    }
  }
  cout << res << '\n';
  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 = 10000 + 233;
int c[maxn];
int n;
bool vis[maxn];
int lowbit(int x) { return x & (-x); }
void add(int x, int d) {
  while (x < maxn) {
    c[x] += d;
    x += lowbit(x);
  }
}
int sum(int x) {
  int ret = 1;
  while (x) {
    ret += c[x];
    x -= lowbit(x);
  }
  return ret;
}
struct node {
  int l, r, p;
} x[maxn], y[maxn];
bool cmpx(node &x, node &y) { return x.p < y.p; }
bool cmpy(node &x, node &y) { return x.r < y.r; }
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  int xcnt = 0, ycnt = 0;
  for (int i = 1; i <= n; i++) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    a += 5001, b += 5001, c += 5001, d += 5001;
    if (a > c) swap(a, c);
    if (b > d) swap(b, d);
    if (a == c) {
      y[++ycnt] = node{b, d, a};
    }
    if (b == d) {
      x[++xcnt] = node{a, c, b};
    }
  }
  sort(x + 1, x + xcnt + 1, cmpx);
  sort(y + 1, y + ycnt + 1, cmpy);
  long long res = 0;
  for (int i = 1; i <= xcnt; i++) {
    memset(vis, 0, sizeof(vis));
    memset(c, 0, sizeof(c));
    for (int j = 1; j <= ycnt; j++) {
      if (y[j].l <= x[i].p && y[j].p >= x[i].l && y[j].p <= x[i].r) {
        vis[j] = 1;
        add(y[j].p, 1);
      }
    }
    int k = 1;
    for (int j = i + 1; j <= xcnt; j++) {
      while (k <= ycnt && y[k].r < x[j].p) {
        if (vis[k]) add(y[k].p, -1);
        k++;
      }
      long long tmp = sum(min(x[i].r, x[j].r)) - sum(max(x[i].l, x[j].l) - 1);
      res += 1ll * tmp * (tmp - 1) / 2;
    }
  }
  cout << res << '\n';
  return 0;
}

```