#include <bits/stdc++.h>
using namespace std;
int n;
struct unu {
  int y, x1, x2;
};
struct doi {
  int x, y1, y2;
};
vector<unu> o;
vector<doi> v;
bool cmp(unu a, unu b) { return a.y < b.y; }
bool cmp2(doi a, doi b) { return a.y2 < b.y2; }
long long ans, nrd;
int aib[100005];
int sum(int x) {
  int ans = 0;
  while (x > 0) {
    ans += aib[x];
    x -= ((x ^ (x - 1)) & x);
  }
  return ans;
}
int query(int l, int r) { return (sum(r) - sum(l - 1)); }
void update(int x, int val) {
  while (x <= 100000) {
    aib[x] += val;
    x += ((x ^ (x - 1)) & x);
  }
}
int main() {
  int i, j, k;
  int x1, y1, x2, y2, l, r;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    x1 += 5001;
    x2 += 5001;
    y1 += 5001;
    y2 += 5001;
    if (x1 == x2)
      v.push_back({x1, min(y1, y2), max(y1, y2)});
    else
      o.push_back({y1, min(x1, x2), max(x1, x2)});
  }
  sort(o.begin(), o.end(), cmp);
  sort(v.begin(), v.end(), cmp2);
  for (i = 0; i < o.size(); i++) {
    memset(aib, 0, sizeof(aib));
    k = v.size() - 1;
    for (j = o.size() - 1; (j > i) && (o[j].y > o[i].y); j--) {
      while ((k >= 0) && (v[k].y2 >= o[j].y)) {
        if (v[k].y1 <= o[i].y) update(v[k].x, 1);
        k--;
      }
      l = max(o[i].x1, o[j].x1);
      r = min(o[i].x2, o[j].x2);
      if (l <= r) {
        nrd = query(l, r);
        ans += (nrd * (nrd - 1) / 2);
      }
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Develop a solution in cpp 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;
int n;
struct unu {
  int y, x1, x2;
};
struct doi {
  int x, y1, y2;
};
vector<unu> o;
vector<doi> v;
bool cmp(unu a, unu b) { return a.y < b.y; }
bool cmp2(doi a, doi b) { return a.y2 < b.y2; }
long long ans, nrd;
int aib[100005];
int sum(int x) {
  int ans = 0;
  while (x > 0) {
    ans += aib[x];
    x -= ((x ^ (x - 1)) & x);
  }
  return ans;
}
int query(int l, int r) { return (sum(r) - sum(l - 1)); }
void update(int x, int val) {
  while (x <= 100000) {
    aib[x] += val;
    x += ((x ^ (x - 1)) & x);
  }
}
int main() {
  int i, j, k;
  int x1, y1, x2, y2, l, r;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    x1 += 5001;
    x2 += 5001;
    y1 += 5001;
    y2 += 5001;
    if (x1 == x2)
      v.push_back({x1, min(y1, y2), max(y1, y2)});
    else
      o.push_back({y1, min(x1, x2), max(x1, x2)});
  }
  sort(o.begin(), o.end(), cmp);
  sort(v.begin(), v.end(), cmp2);
  for (i = 0; i < o.size(); i++) {
    memset(aib, 0, sizeof(aib));
    k = v.size() - 1;
    for (j = o.size() - 1; (j > i) && (o[j].y > o[i].y); j--) {
      while ((k >= 0) && (v[k].y2 >= o[j].y)) {
        if (v[k].y1 <= o[i].y) update(v[k].x, 1);
        k--;
      }
      l = max(o[i].x1, o[j].x1);
      r = min(o[i].x2, o[j].x2);
      if (l <= r) {
        nrd = query(l, r);
        ans += (nrd * (nrd - 1) / 2);
      }
    }
  }
  cout << ans;
  return 0;
}

```