E. Two Teams

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E. Two Teams

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are nn students standing in a row. Two coaches are forming two teams — the first coach chooses the first team and the second coach chooses the second team.

The ii-th student has integer programming skill aiai. All programming skills are distinct and between 11 and nn, inclusive.

Firstly, the first coach will choose the student with maximum programming skill among all students not taken into any team, and kk closest students to the left of him and kk closest students to the right of him (if there are less than kk students to the left or to the right, all of them will be chosen). All students that are chosen leave the row and join the first team. Secondly, the second coach will make the same move (but all students chosen by him join the second team). Then again the first coach will make such move, and so on. This repeats until the row becomes empty (i. e. the process ends when each student becomes to some team).

Your problem is to determine which students will be taken into the first team and which students will be taken into the second team.

Input

The first line of the input contains two integers nn and kk (1≤k≤n≤2⋅1051≤k≤n≤2⋅105) — the number of students and the value determining the range of chosen students during each move, respectively.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤n1≤ai≤n), where aiai is the programming skill of the ii-th student. It is guaranteed that all programming skills are distinct.

Output

Print a string of nn characters; ii-th character should be 1 if ii-th student joins the first team, or 2 otherwise.

Examples

input

Copy

5 2
2 4 5 3 1

output

Copy

input

Copy

5 1
2 1 3 5 4

output

Copy

input

Copy

7 1
7 2 1 3 5 4 6

output

Copy

input

Copy

5 1
2 4 5 3 1

output

Copy

Note

In the first example the first coach chooses the student on a position 33, and the row becomes empty (all students join the first team).

In the second example the first coach chooses the student on position 44, and the row becomes [2,1][2,1] (students with programming skills [3,4,5][3,4,5] join the first team). Then the second coach chooses the student on position 11, and the row becomes empty (and students with programming skills [1,2][1,2] join the second team).

In the third example the first coach chooses the student on position 11, and the row becomes [1,3,5,4,6][1,3,5,4,6] (students with programming skills [2,7][2,7] join the first team). Then the second coach chooses the student on position 55, and the row becomes [1,3,5][1,3,5] (students with programming skills [4,6][4,6] join the second team). Then the first coach chooses the student on position 33, and the row becomes [1][1] (students with programming skills [3,5][3,5] join the first team). And then the second coach chooses the remaining student (and the student with programming skill 11 joins the second team).

In the fourth example the first coach chooses the student on position 33, and the row becomes [2,1][2,1] (students with programming skills [3,4,5][3,4,5] join the first team). Then the second coach chooses the student on position 11, and the row becomes empty (and students with programming skills [1,2][1,2] join the second team).

题意：

有n名学生站成一排。两位教练组成了两支球队 - 第一支教练选择了第一支球队，第二支教练选择了第二支球队。

第i名学生具有整数编程技能ai。所有编程技能都是不同的，介于1和n之间。

首先，第一位教练将选择所有没有进入任何团队的学生中具有最高编程技能的学生，以及在他左边的k个最近的学生和在他右边的k个最近的学生（如果有少于k个学生到向左或向右，将选择所有这些。所有被选中的学生都会离开队伍并加入第一队。其次，第二个教练将采取同样的行动（但他选择的所有学生都加入了第二个团队）。然后第一位教练将再次进行这样的行动，依此类推。这将重复直到该行变空（即，当每个学生成为某个团队时，该过程结束）。

你的问题是确定哪些学生将被带入第一队，哪些学生将被带入第二队。

方法：区间删除

#include<bits/stdc++.h>
using namespace std;
const int maxn=2e5+10;
typedef long long LL;
const LL mod=1e9+7;
int num[200010],ans[200010];
int pos[200010];//标记位置
int L[200010],R[200010];//用于区间删除
int vis[200010];//标记
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&num[i]);
        vis[i]=0;
        ans[i]=num[i];//复制
        pos[num[i]]=i;//id
        L[i]=i-1;//连接
        R[i]=i+1;
    }
    sort(ans+1,ans+1+n);
    int in=1;
    for(int i=n;i>=1;i--)
    {
        int item=pos[ans[i]];
        if(vis[item]) continue;//标记
            vis[item]=in;
        int t=k;
        int l=L[item];
        while(t--&&l>=1){
            vis[l]=in;
            l--;
        }
        int r=R[item];
        t=k;
        while(t--&&r<=n){
            vis[r]=in;
            r=R[r];
        }
        L[r]=l;
        R[l]=r;//缩减去除中间区间
        if(in==1) in=2;
        else in=1;
    }

    for(int i=1;i<=n;i++)
        printf("%d",vis[i]);
    return 0;
}