#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll A,B;
int n,m;
const int N=500010;
int a[N];
int ans=1e9+7;
int nn;
inline int check(int L){
	int ret=0;
	for(int i=1;i<=L;i++){
		if(a[i]+a[L-i+1]>=m){
			return -1;
		}
		ret=max(ret,a[i]+a[L-i+1]);
	}
	for(int i=L+1;i<=n;i++){
		int rr=i-L;rr=n-rr+1;
		if(a[i]+a[rr]<m)return -2;
		ret=max(ret,(a[i]+a[rr])%m);
	}
	ans = ret;
	return 1;
}

int main()
{
	cin>>nn>>m;
	n=nn*2;
	for(int i=1;i<=n;i++)scanf("%d",&a[i]);
	sort(a+1,a+n+1);
	int l=0,r=nn;
	while(l<r){
		int mid=(l+r)>>1;
		if(check(mid*2)==-2) l=mid+1;
		else r=mid;
	}
	check(l*2);
	cout<<ans<<endl;
}

### Prompt

In Cpp, your task is to solve the following problem:

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.

Examples

Input

3 10
0 2 3 4 5 9


Output

5


Input

2 10
1 9 1 9


Output

0

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll A,B;
int n,m;
const int N=500010;
int a[N];
int ans=1e9+7;
int nn;
inline int check(int L){
	int ret=0;
	for(int i=1;i<=L;i++){
		if(a[i]+a[L-i+1]>=m){
			return -1;
		}
		ret=max(ret,a[i]+a[L-i+1]);
	}
	for(int i=L+1;i<=n;i++){
		int rr=i-L;rr=n-rr+1;
		if(a[i]+a[rr]<m)return -2;
		ret=max(ret,(a[i]+a[rr])%m);
	}
	ans = ret;
	return 1;
}

int main()
{
	cin>>nn>>m;
	n=nn*2;
	for(int i=1;i<=n;i++)scanf("%d",&a[i]);
	sort(a+1,a+n+1);
	int l=0,r=nn;
	while(l<r){
		int mid=(l+r)>>1;
		if(check(mid*2)==-2) l=mid+1;
		else r=mid;
	}
	check(l*2);
	cout<<ans<<endl;
}

```