#include <bits/stdc++.h>
using namespace std;
const int MX = 1e4 + 1;
vector<tuple<int, int, int> > ver, hor;
vector<int> mst[MX * 2];
void build() {
  for (int t = MX - 1; t; t--) {
    auto &v = mst[t], v1 = mst[2 * t], v2 = mst[2 * t + 1];
    int i = 0, j = 0;
    while (i < v1.size() && j < v2.size())
      v.push_back(v1[i] < v2[j] ? v1[i++] : v2[j++]);
    while (i < v1.size()) v.push_back(v1[i++]);
    while (j < v2.size()) v.push_back(v2[j++]);
  }
}
int count_ge(vector<int>& v, int x) {
  return v.size() - (lower_bound(v.begin(), v.end(), x) - v.begin());
}
int qry(int x, int y1, int y2) {
  y1 += MX, y2 += MX + 1;
  int ret = 0;
  while (y1 < y2) {
    if (y1 & 1) ret += count_ge(mst[y1++], x);
    if (y2 & 1) ret += count_ge(mst[--y2], x);
    y1 /= 2, y2 /= 2;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += 5000;
    y1 += 5000;
    x2 += 5000;
    y2 += 5000;
    if (x1 == x2) {
      ver.push_back({x1, min(y1, y2), max(y1, y2)});
    } else {
      hor.push_back({y1, min(x1, x2), max(x1, x2)});
    }
  }
  sort(ver.begin(), ver.end(),
       [](tuple<int, int, int> a, tuple<int, int, int> b) {
         return get<0>(a) < get<0>(b);
       });
  int vsz = ver.size(), hsz = hor.size(), j = 0;
  long long ans = 0;
  for (int i = 0; i < vsz; i = j) {
    int x = get<0>(ver[i]);
    while (j < vsz && x == get<0>(ver[j])) {
      j++;
    }
    for (int k = 0; k < MX * 2; k++) {
      mst[k].clear();
    }
    for (int k = 0; k < hsz; k++) {
      int hy, hx1, hx2;
      tie(hy, hx1, hx2) = hor[k];
      if (hx1 <= x && x <= hx2) mst[MX + hy] = {hx2};
    }
    build();
    for (int t = i; t < j; t++) {
      int x, y1, y2;
      tie(x, y1, y2) = ver[t];
      for (int k = j; k < vsz; k++) {
        int vx, vy1, vy2;
        tie(vx, vy1, vy2) = ver[k];
        int c = qry(vx, max(y1, vy1), min(y2, vy2));
        ans += c * (c - 1) / 2;
      }
    }
  }
  cout << ans;
}

### Prompt

Please formulate a cpp solution to 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 MX = 1e4 + 1;
vector<tuple<int, int, int> > ver, hor;
vector<int> mst[MX * 2];
void build() {
  for (int t = MX - 1; t; t--) {
    auto &v = mst[t], v1 = mst[2 * t], v2 = mst[2 * t + 1];
    int i = 0, j = 0;
    while (i < v1.size() && j < v2.size())
      v.push_back(v1[i] < v2[j] ? v1[i++] : v2[j++]);
    while (i < v1.size()) v.push_back(v1[i++]);
    while (j < v2.size()) v.push_back(v2[j++]);
  }
}
int count_ge(vector<int>& v, int x) {
  return v.size() - (lower_bound(v.begin(), v.end(), x) - v.begin());
}
int qry(int x, int y1, int y2) {
  y1 += MX, y2 += MX + 1;
  int ret = 0;
  while (y1 < y2) {
    if (y1 & 1) ret += count_ge(mst[y1++], x);
    if (y2 & 1) ret += count_ge(mst[--y2], x);
    y1 /= 2, y2 /= 2;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x1, y1, x2, y2;
    cin >> x1 >> y1 >> x2 >> y2;
    x1 += 5000;
    y1 += 5000;
    x2 += 5000;
    y2 += 5000;
    if (x1 == x2) {
      ver.push_back({x1, min(y1, y2), max(y1, y2)});
    } else {
      hor.push_back({y1, min(x1, x2), max(x1, x2)});
    }
  }
  sort(ver.begin(), ver.end(),
       [](tuple<int, int, int> a, tuple<int, int, int> b) {
         return get<0>(a) < get<0>(b);
       });
  int vsz = ver.size(), hsz = hor.size(), j = 0;
  long long ans = 0;
  for (int i = 0; i < vsz; i = j) {
    int x = get<0>(ver[i]);
    while (j < vsz && x == get<0>(ver[j])) {
      j++;
    }
    for (int k = 0; k < MX * 2; k++) {
      mst[k].clear();
    }
    for (int k = 0; k < hsz; k++) {
      int hy, hx1, hx2;
      tie(hy, hx1, hx2) = hor[k];
      if (hx1 <= x && x <= hx2) mst[MX + hy] = {hx2};
    }
    build();
    for (int t = i; t < j; t++) {
      int x, y1, y2;
      tie(x, y1, y2) = ver[t];
      for (int k = j; k < vsz; k++) {
        int vx, vy1, vy2;
        tie(vx, vy1, vy2) = ver[k];
        int c = qry(vx, max(y1, vy1), min(y2, vy2));
        ans += c * (c - 1) / 2;
      }
    }
  }
  cout << ans;
}

```