#include <bits/stdc++.h>
using namespace std;
const int MX = 10000;
const int inf = 1e9;
const long long mod = 998244353;
const int ad = 5001;
long long T[4 * MX];
struct segment {
  int x1, y1, x2, y2;
};
bool comp1(segment &a, segment &b) { return a.y2 < b.y2; }
bool comp2(segment &a, segment &b) { return a.y2 < b.y2; }
void up(int p, int l, int h, int x, int vl) {
  if (l == h) {
    T[p] += vl;
    return;
  }
  int m = (l + h) / 2;
  if (x <= m)
    up(2 * p, l, m, x, vl);
  else
    up(2 * p + 1, m + 1, h, x, vl);
  T[p] = T[2 * p] + T[2 * p + 1];
}
long long Q(int p, int l, int h, int x, int y) {
  if (h < x || l > y) return 0;
  if (l >= x && h <= y) return T[p];
  int m = (l + h) / 2;
  return Q(2 * p, l, m, x, y) + Q(2 * p + 1, m + 1, h, x, y);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  vector<segment> ver, hor;
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += ad;
    y1 += ad;
    x2 += ad;
    y2 += ad;
    if (y1 == y2) {
      if (x1 > x2) swap(x1, x2);
      hor.push_back({x1, y1, x2, y2});
    } else {
      if (y1 > y2) swap(y1, y2);
      ver.push_back({x1, y1, x2, y2});
    }
  }
  sort(hor.begin(), hor.end(), comp1);
  sort(ver.begin(), ver.end(), comp2);
  long long sm = 0;
  for (int i = 0; i < hor.size(); i++) {
    vector<pair<long long, long long> > good;
    for (int j = ver.size() - 1; j >= 0; j--) {
      if (ver[j].y1 <= hor[i].y1 && ver[j].y2 >= hor[i].y1 &&
          ver[j].x1 >= hor[i].x1 && ver[j].x1 <= hor[i].x2) {
        good.push_back({ver[j].x1, ver[j].y2});
        up(1, 1, MX, ver[j].x1, 1);
      }
    }
    for (int j = i + 1; j < hor.size(); j++) {
      if (hor[i].y1 >= hor[j].y1) continue;
      int l = max(hor[i].x1, hor[j].x1);
      int r = min(hor[i].x2, hor[j].x2);
      if (l > r) continue;
      while (good.size() && good.back().second < hor[j].y1) {
        int id = good.back().first;
        good.pop_back();
        up(1, 1, MX, id, -1);
      }
      long long cnt = Q(1, 1, MX, l, r);
      sm += (cnt * (cnt - 1)) / 2;
    }
    while (good.size()) {
      int id = good.back().first;
      good.pop_back();
      up(1, 1, MX, id, -1);
    }
  }
  cout << sm << endl;
  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;
const int MX = 10000;
const int inf = 1e9;
const long long mod = 998244353;
const int ad = 5001;
long long T[4 * MX];
struct segment {
  int x1, y1, x2, y2;
};
bool comp1(segment &a, segment &b) { return a.y2 < b.y2; }
bool comp2(segment &a, segment &b) { return a.y2 < b.y2; }
void up(int p, int l, int h, int x, int vl) {
  if (l == h) {
    T[p] += vl;
    return;
  }
  int m = (l + h) / 2;
  if (x <= m)
    up(2 * p, l, m, x, vl);
  else
    up(2 * p + 1, m + 1, h, x, vl);
  T[p] = T[2 * p] + T[2 * p + 1];
}
long long Q(int p, int l, int h, int x, int y) {
  if (h < x || l > y) return 0;
  if (l >= x && h <= y) return T[p];
  int m = (l + h) / 2;
  return Q(2 * p, l, m, x, y) + Q(2 * p + 1, m + 1, h, x, y);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  vector<segment> ver, hor;
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += ad;
    y1 += ad;
    x2 += ad;
    y2 += ad;
    if (y1 == y2) {
      if (x1 > x2) swap(x1, x2);
      hor.push_back({x1, y1, x2, y2});
    } else {
      if (y1 > y2) swap(y1, y2);
      ver.push_back({x1, y1, x2, y2});
    }
  }
  sort(hor.begin(), hor.end(), comp1);
  sort(ver.begin(), ver.end(), comp2);
  long long sm = 0;
  for (int i = 0; i < hor.size(); i++) {
    vector<pair<long long, long long> > good;
    for (int j = ver.size() - 1; j >= 0; j--) {
      if (ver[j].y1 <= hor[i].y1 && ver[j].y2 >= hor[i].y1 &&
          ver[j].x1 >= hor[i].x1 && ver[j].x1 <= hor[i].x2) {
        good.push_back({ver[j].x1, ver[j].y2});
        up(1, 1, MX, ver[j].x1, 1);
      }
    }
    for (int j = i + 1; j < hor.size(); j++) {
      if (hor[i].y1 >= hor[j].y1) continue;
      int l = max(hor[i].x1, hor[j].x1);
      int r = min(hor[i].x2, hor[j].x2);
      if (l > r) continue;
      while (good.size() && good.back().second < hor[j].y1) {
        int id = good.back().first;
        good.pop_back();
        up(1, 1, MX, id, -1);
      }
      long long cnt = Q(1, 1, MX, l, r);
      sm += (cnt * (cnt - 1)) / 2;
    }
    while (good.size()) {
      int id = good.back().first;
      good.pop_back();
      up(1, 1, MX, id, -1);
    }
  }
  cout << sm << endl;
  return 0;
}

```