Mice and Rice is the name of a programming contest in which each programmer must write a piece of code to control the movements of a mouse in a given map. The goal of each mouse is to eat as much rice as possible in order to become a FatMouse.

First the playing order is randomly decided for N_P programmers. Then every N_G programmers are grouped in a match. The fattest mouse in a group wins and enters the next turn. All the losers in this turn are ranked the same. Every N_G winners are then grouped in the next match until a final winner is determined.

For the sake of simplicity, assume that the weight of each mouse is fixed once the programmer submits his/her code. Given the weights of all the mice and the initial playing order, you are supposed to output the ranks for the programmers.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers: N_P and N_G (<= 1000), the number of programmers and the maximum number of mice in a group, respectively. If there are less than N_G mice at the end of the player's list, then all the mice left will be put into the last group. The second line contains N_P distinct non-negative numbers W_i (i=0,...N_P-1) where each W_i is the weight of the i-th mouse respectively. The third line gives the initial playing order which is a permutation of 0,...N_P-1 (assume that the programmers are numbered from 0 to N_P-1). All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the final ranks in a line. The i-th number is the rank of the i-th programmer, and all the numbers must be separated by a space, with no extra space at the end of the line.

Sample Input:

11 3
25 18 0 46 37 3 19 22 57 56 10
6 0 8 7 10 5 9 1 4 2 3

Sample Output:

5 5 5 2 5 5 5 3 1 3 5

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<queue>
 6 #include<vector>
 7 #include<cmath>
 8 #include<string>
 9 #include<map>
10 #include<set>
11 using namespace std;
12 int ra[1005],weight[1005],order[1005];
13 int main(){
14     //freopen("D:\\INPUT.txt","r",stdin);
15     int np,ng;
16     scanf("%d %d",&np,&ng);
17     int i,j;
18     for(i=0;i<np;i++){
19         scanf("%d",&weight[i]);
20     }
21     for(i=0;i<np;i++){
22         scanf("%d",&order[i]);
23     }
24
25
26     int nnp=np;
27     int r=nnp/ng+(nnp%ng==0?0:1)+1;//每局的等级;
28     int s,maxnum;
29     queue<int> q;//存放编号
30     for(i=0;i<nnp;){
31         ra[order[i]]=r;
32         maxnum=order[i];//每一轮都有最大值
33         s=i++;
34         for(;i<nnp&&i-s<ng;i++){
35             ra[order[i]]=r;
36             if(weight[maxnum]<weight[order[i]]){
37                 maxnum=order[i];
38             }
39         }
40         q.push(maxnum);//每组的最大值进入到下一次比赛中
41         //cout<<maxnum<<endl;
42     }
43     while(q.size()>1){
44         nnp=q.size();
45         r=nnp/ng+(nnp%ng==0?0:1)+1;
46         for(i=0;i<nnp;){
47             maxnum=q.front();
48             q.pop();
49             ra[maxnum]=r;
50             s=i++;
51             for(;i<nnp&&i-s<ng;i++){
52                 ra[q.front()]=r;
53                 if(weight[maxnum]<weight[q.front()]){
54                     maxnum=q.front();
55                 }
56                 q.pop();//不要忘
57             }
58             q.push(maxnum);
59         }
60     }
61     ra[q.front()]=1;
62     printf("%d",ra[0]);
63     for(i=1;i<np;i++){
64         printf(" %d",ra[i]);
65     }
66     return 0;
67 }