#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
long long POW(long long a, long long b, long long MMM = MOD) {
  long long ret = 1;
  for (; b; b >>= 1, a = (a * a) % MMM)
    if (b & 1) ret = (ret * a) % MMM;
  return ret;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) {
  if (a == 0 || b == 0) return a + b;
  return a * (b / gcd(a, b));
}
int dx[] = {0, 1, 0, -1, 1, 1, -1, -1}, dy[] = {1, 0, -1, 0, 1, -1, 1, -1};
int ddx[] = {2, 2, -2, -2, 1, 1, -1, -1}, ddy[] = {1, -1, 1, -1, 2, -2, 2, -2};
vector<pair<int, int> > m[10005], l[10005];
vector<int> v[10005];
int tree[10005];
int n;
void upd(int i, int k) {
  while (i <= 10001) {
    tree[i] += k;
    i += (i & -i);
  }
}
int sum(int i) {
  int c = 0;
  while (i > 0) {
    c += tree[i];
    i -= (i & -i);
  }
  return c;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int(i) = (0); (i) <= (n - 1); (i) += (1)) {
    int x1, y22, x2, y2;
    scanf("%d%d%d%d", &x1, &y22, &x2, &y2);
    x1 += 5001, x2 += 5001, y22 += 5001, y2 += 5001;
    if (x1 == x2)
      l[x1].push_back(make_pair(min(y22, y2), max(y22, y2)));
    else
      m[y22].push_back(make_pair(min(x1, x2), max(x1, x2)));
  }
  long long ans = 0;
  for (int(y) = (1); (y) <= (10001); (y) += (1)) {
    for (pair<int, int> p : m[y]) {
      for (int(i) = (1); (i) <= (10001); (i) += (1)) v[i].clear();
      memset((tree), 0, sizeof(tree));
      ;
      for (int(x) = (p.first); (x) <= (p.second); (x) += (1)) {
        for (pair<int, int> p2 : l[x]) {
          if (p2.first <= y && p2.second > y) {
            v[p2.second].push_back(x);
            upd(x, 1);
          }
        }
      }
      for (int(y2) = (y + 1); (y2) <= (10001); (y2) += (1)) {
        for (pair<int, int> p2 : m[y2]) {
          int t = sum(p2.second) - sum(p2.first - 1);
          ans += t * (t - 1) / 2;
        }
        for (int x : v[y2]) upd(x, -1);
      }
    }
  }
  printf("%lld", ans);
}

### 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 MOD = 1000000007;
long long POW(long long a, long long b, long long MMM = MOD) {
  long long ret = 1;
  for (; b; b >>= 1, a = (a * a) % MMM)
    if (b & 1) ret = (ret * a) % MMM;
  return ret;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) {
  if (a == 0 || b == 0) return a + b;
  return a * (b / gcd(a, b));
}
int dx[] = {0, 1, 0, -1, 1, 1, -1, -1}, dy[] = {1, 0, -1, 0, 1, -1, 1, -1};
int ddx[] = {2, 2, -2, -2, 1, 1, -1, -1}, ddy[] = {1, -1, 1, -1, 2, -2, 2, -2};
vector<pair<int, int> > m[10005], l[10005];
vector<int> v[10005];
int tree[10005];
int n;
void upd(int i, int k) {
  while (i <= 10001) {
    tree[i] += k;
    i += (i & -i);
  }
}
int sum(int i) {
  int c = 0;
  while (i > 0) {
    c += tree[i];
    i -= (i & -i);
  }
  return c;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int(i) = (0); (i) <= (n - 1); (i) += (1)) {
    int x1, y22, x2, y2;
    scanf("%d%d%d%d", &x1, &y22, &x2, &y2);
    x1 += 5001, x2 += 5001, y22 += 5001, y2 += 5001;
    if (x1 == x2)
      l[x1].push_back(make_pair(min(y22, y2), max(y22, y2)));
    else
      m[y22].push_back(make_pair(min(x1, x2), max(x1, x2)));
  }
  long long ans = 0;
  for (int(y) = (1); (y) <= (10001); (y) += (1)) {
    for (pair<int, int> p : m[y]) {
      for (int(i) = (1); (i) <= (10001); (i) += (1)) v[i].clear();
      memset((tree), 0, sizeof(tree));
      ;
      for (int(x) = (p.first); (x) <= (p.second); (x) += (1)) {
        for (pair<int, int> p2 : l[x]) {
          if (p2.first <= y && p2.second > y) {
            v[p2.second].push_back(x);
            upd(x, 1);
          }
        }
      }
      for (int(y2) = (y + 1); (y2) <= (10001); (y2) += (1)) {
        for (pair<int, int> p2 : m[y2]) {
          int t = sum(p2.second) - sum(p2.first - 1);
          ans += t * (t - 1) / 2;
        }
        for (int x : v[y2]) upd(x, -1);
      }
    }
  }
  printf("%lld", ans);
}

```