Problem Description

This problem is the most boring one you've ever seen.

Given a sequence of integers a1,?a2,?...,?an and a non-negative integer h, our goal is to partition the sequence into two subsequences (not necessarily consist of continuous elements). Each element of the original sequence should be contained in exactly one of the result subsequences. Note, that one of the result subsequences can be empty.

Let's define function f(ai,?aj) on pairs of distinct elements (that is i?≠?j) in the original sequence. If ai and aj are in the same subsequence in the current partition then f(ai,?aj)?=?ai?+?aj otherwise f(ai,?aj)?=?ai?+?aj?+?h.

Consider all possible values of the function f for some partition. We'll call the goodness of this partiotion the difference between the maximum value of function f and the minimum value of function f.

Your task is to find a partition of the given sequence a that have the minimal possible goodness among all possible partitions.

Input

The first line of input contains integers n and h (2 ≤ n ≤ 105, 0 ≤ h ≤ 108). In the second line there is a list of n space-separated integers representing a1, a2, ..., an (0 ≤ ai ≤ 108).

Output

The first line of output should contain the required minimum goodness.

The second line describes the optimal partition. You should print n whitespace-separated integers in the second line. The i-th integer is 1 if ai is in the first subsequence otherwise it should be 2.

If there are several possible correct answers you are allowed to print any of them.

Examples

Input

3 2
1 2 3

Output

1
1 2 2?

Input

5 10
0 1 0 2 1

Output

3
2 2 2 2 2?

Note

In the first sample the values of f are as follows: f(1,?2)?=?1?+?2?+?2?=?5, f(1,?3)?=?1?+?3?+?2?=?6 and f(2,?3)?=?2?+?3?=?5. So the difference between maximum and minimum values of f is 1.

In the second sample the value of h is large, so it's better for one of the sub-sequences to be empty.

題意：給出 n 個數和一個數 h，現在要將 n 個數分成兩個序列，定義一個函數 f(ai,aj)，當 ai、aj 處于同一序列時，f(ai,aj)=ai+a[j，當 ai、aj 處于不同序列時，f(ai,aj)=ai+aj+h，現要使得函數 f 的最大值與最小值的差最小，求最小值

思路：

要使得最大值與最小值的差最小，那么就要讓最大值盡可能的小，最小值盡可能的大

首先將這 n 個數從小到大進行排序，在排序后可以發現，若將大的數移動到另一序列，只能使得最大值更大，無法使最小值更大，而將最小的數移動到另一序列，不僅使得最大值變小了，也使得最小值變大了

那么問題一共有兩種分配方案：

• 當只有一個序列時
• 一個集合序列中為最小的元素 min 時，一個集合序列為剩下的元素時

因此比較兩種分配方案哪個的差最小即可

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #include<bitset> #define EPS 1e-9 #define PI acos(-1.0) #define INF 0x3f3f3f3f #define LL long long const int MOD = 1E9+7; const int N = 1000000+5; const int dx[] = {-1,1,0,0,-1,-1,1,1}; const int dy[] = {0,0,-1,1,-1,1,-1,1}; using namespace std;struct Node{int val;int pos;Node(){}Node(int val,int pos):val(val),pos(pos){}bool operator < (const Node &rhs)const{return val<rhs.val;} }node[N]; int main() {int n,h;scanf("%d%d",&n,&h);for(int i=1;i<=n;i++){scanf("%d",&node[i].val);node[i].pos=i;}sort(node+1,node+1+n);int sameMaxx=node[n].val+node[n-1].val;int sameMinn=node[1].val+node[2].val;int sameSub=sameMaxx-sameMinn;int unsameMaxx=max(node[n].val+node[1].val+h,node[n].val+node[n-1].val);int unsameMinn=min(node[1].val+node[2].val+h,node[2].val+node[3].val);int unsameSub=unsameMaxx-unsameMinn;int minn=min(abs(sameSub),abs(unsameSub));printf("%d\n",minn);int pos=node[1].pos;for(int i=1;i<=n;i++){if(i==pos&&abs(unsameSub)<abs(sameSub))printf("1 ");elseprintf("2 ");}printf("\n");return 0; }

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