Problem:
Score : 1200 points

Problem Statement

Let M be a positive integer.

You are given 2 N integers a_1, a_2, \ldots, a_{2 N}, where 0 \leq a_i < M for each i.

Consider dividing the 2 N integers into N pairs.
Here, each integer must belong to exactly one pair.

We define the ugliness of a pair (x, y) as (x + y) \mod M.
Let Z be the largest ugliness of the N pairs. Find the minimum possible value of Z.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq M \leq 10^9

0 \leq a_i < M

Input

Input is given from Standard Input in the following format:

N M
a_1 a_2 \cdots a_{2N}

Output

Print the minimum possible value of Z, where Z is the largest ugliness of the N pairs.

Sample Input 1

3 10
0 2 3 4 5 9

Sample Output 1

5

One solution is to form pairs (0, 5), (2, 3) and (4, 9), with ugliness 5, 5 and 3, respectively.

Sample Input 2

2 10
1 9 1 9

Sample Output 2

0

Pairs (1, 9) and (1, 9) should be formed, with ugliness both 0.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<cmath>
#include<stack>
#include<algorithm>
#define rg register
#define ll long long
#define LDB long double
#define ull unsigned long long
#define view(i,x) for(rg int i=hd[x];i!=-1;i=e[i].nt)
#define go(i,x,a) for(rg int i=a;i<x;i++)
#define inf 0x3f3f3f3f
#define INF 0x7fffffff
using namespace std;

const int maxn=2*1e5+5;
int n,m,a[maxn];

inline int rd(){
	int ret=0,af=1; char gc=getchar();
	while(gc < '0' || gc > '9'){ if(gc=='-') af=-af; gc=getchar(); }
	while(gc >= '0' && gc <= '9') ret=ret*10+gc-'0',gc=getchar();
	return ret*af;
}

inline bool check(int x){
	go(i,x+1,1)
		if(a[i]+a[2*x-i+1] >= m) return 0;
	go(i,n*2+1,x*2+1){
		if(i > 2*n-(i-x*2)+1) break;
		if(a[i] + a[2*n-(i-x*2)+1] < m) return 0;
	}
	return 1;
}

inline void gmax(int &a,int b){
	if(a < b) a=b; return ;
}

int main(){
	#ifndef ONLINE_JUDGE
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	#endif
	n=rd(); m=rd();
	go(i,n*2+1,1) a[i]=rd();
	sort(a+1,a+2*n+1);
	int l=0,r=n,res=-1;
	while(l <= r){
		int mid=l+r>>1;
		if(check(mid)) res=mid,r=mid-1;
		else l=mid+1;
	}
	int ans=0;
	int x=res;
	go(i,x+1,1)
		gmax(ans,a[i]+a[2*x-i+1]);
	go(i,n*2+1,x*2+1){
		if(i > 2*n-(i-x*2)+1) break;
		gmax(ans,(a[i]+a[2*n-(i-x*2)+1])%m);
	}
	printf("%d",ans);
	return 0;
}//Faze