Educational Round 27

A. Chess Tourney

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Berland annual chess tournament is coming!

Organizers have gathered 2·n chess players who should be divided into two teams with n people each. The first team is sponsored by BerOil and the second team is sponsored by BerMobile. Obviously, organizers should guarantee the win for the team of BerOil.

Thus, organizers should divide all 2·n players into two teams with n people each in such a way that the first team always wins.

Every chess player has its rating r_i. It is known that chess player with the greater rating always wins the player with the lower rating. If their ratings are equal then any of the players can win.

After teams assignment there will come a drawing to form n pairs of opponents: in each pair there is a player from the first team and a player from the second team. Every chess player should be in exactly one pair. Every pair plays once. The drawing is totally random.

Is it possible to divide all 2·n players into two teams with n people each so that the player from the first team in every pair wins regardless of the results of the drawing?

 

Input

The first line contains one integer n (1 ≤ n ≤ 100).

The second line contains 2·n integers a₁, a₂, ... a_2n (1 ≤ a_i ≤ 1000).

 

Output

If it's possible to divide all 2·n players into two teams with n people each so that the player from the first team in every pair wins regardless of the results of the drawing, then print "YES". Otherwise print "NO".

 

Examples

 

Input

2
1 3 2 4

Output

YES

 

Input

1
3 3

Output

NO

题意：把2n个人分成两队，每个数代表一个等级，等级高的胜等级低的，如果等级相等，怎不知道的谁胜谁负，
两队里的每个人都有可能比赛，所以要胜利的话，就要大于，
分析数据，在sort后，实际只要判断中间两项就好，这是由于从中间截成两队的原因

#include <bits/stdc++.h>
using namespace std;
int a[205];
int main(){
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n+n;i++)
        scanf("%d",&a[i]);
    sort(a+1,a+1+(n<<1));
    if(a[n]<a[n+1]) printf("YES
");//a[5]<a[6]
    else printf("NO
");
    return 0;
}