AIM Tech Round 5C. Rectangles 思维

C. Rectangles

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given nn rectangles on a plane with coordinates of their bottom left and upper right points. Some (n−1)(n−1) of the given nn rectangles have some common point. A point belongs to a rectangle if this point is strictly inside the rectangle or belongs to its boundary.

Find any point with integer coordinates that belongs to at least (n−1)(n−1) given rectangles.

Input

The first line contains a single integer nn (2≤n≤1326742≤n≤132674) — the number of given rectangles.

Each the next nn lines contains four integers x1x1, y1y1, x2x2 and y2y2 (−109≤x1<x2≤109−109≤x1<x2≤109, −109≤y1<y2≤109−109≤y1<y2≤109) — the coordinates of the bottom left and upper right corners of a rectangle.

Output

Print two integers xx and yy — the coordinates of any point that belongs to at least (n−1)(n−1) given rectangles.

Examples

input

Copy

3
0 0 1 1
1 1 2 2
3 0 4 1

output

Copy

1 1

input

Copy

3
0 0 1 1
0 1 1 2
1 0 2 1

output

Copy

1 1

input

Copy

4
0 0 5 5
0 0 4 4
1 1 4 4
1 1 4 4

output

Copy

1 1

input

Copy

5
0 0 10 8
1 2 6 7
2 3 5 6
3 4 4 5
8 1 9 2

output

Copy

3 4

Note

The picture below shows the rectangles in the first and second samples. The possible answers are highlighted.

题意：给出n个矩形，找一个点至少同时在n-1个矩形内。

思路：我们分别对每条对角线求前缀交和后缀交，则若在每个条对角线左右两边的的前缀与后缀取交后还存在交点，即为解。

代码：

 #include"bits/stdc++.h"

 #define db double
 #define ll long long
 #define vl vector<ll>
 #define ci(x) scanf("%d",&x)
 #define cd(x) scanf("%lf",&x)
 #define cl(x) scanf("%lld",&x)
 #define pi(x) printf("%d\n",x)
 #define pd(x) printf("%f\n",x)
 #define pl(x) printf("%lld\n",x)
 #define rep(i, n) for(int i=0;i<n;i++)
 using namespace std;
 const int N = 1e6 + ;
 const int mod = 1e9 + ;
 const int MOD = ;
 const db  PI = acos(-1.0);
 const db  eps = 1e-;
 const ll  INF = 0x3fffffffffffffff;
 int n;
 struct P{
     int d,l,u,r;
     inline P operator | (P a){
         return (P){max(a.d,d),max(a.l,l),min(a.u,u),min(a.r,r)};
     }
 }a[N],pre[N],suf[N];
 int main(){
     ci(n);
     for(int i=;i<=n;i++){
        ci(a[i].d),ci(a[i].l),ci(a[i].u),ci(a[i].r);
     }
     pre[]=suf[n+]={-mod,-mod,mod,mod};//初始化
     for(int i=;i<=n;i++){
         pre[i]=pre[i-]|a[i];//前缀
     }
     for(int i=n;i>=;i--){
         suf[i]=suf[i+]|a[i];//后缀
     }
     for(int i=;i<=n;i++){
         P tmp=pre[i-]|suf[i+];//取交
         if(tmp.d<=tmp.u&&tmp.l<=tmp.r) return !printf("%d %d\n",tmp.d,tmp.l);
     }
     return ;
 }