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: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define inf 1e18

ll m,ansmax,a[200009];

int n;

bool check(int pos)
{
    int i,j;
    ll ans=0LL;
    for(i=pos,j=n*2-1;i<j;i++,j--)
    {
        if(a[i]+a[j]<m) return false;
        ans=max(ans,a[i]+a[j]-m);
    }
    for(i=0,j=pos-1;i<j;i++,j--)
        ans=max(ans,a[i]+a[j]);
    ansmax=min(ans,ansmax);
    return true;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int i,j,lo,hi,mid;
    cin >> n >> m;
    for(i=0;i<2*n;i++)
        cin >> a[i];
    sort(a,a+2*n);

    lo=0;
    hi=n;
    ansmax=inf;
    while(lo<hi)
    {
        mid=(lo+hi)/2;
        if(check(mid*2))
            hi=mid;
        else lo=mid+1;

    }
    check(lo*2);
    check(hi*2);
    cout << ansmax << endl;
}