1442

/*
树状数组，二分找第K个
*/

// include file
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <cctype>
#include <ctime>

#include <iostream>
#include <sstream>
#include <fstream>
#include <iomanip>
#include <bitset>

#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <list>
#include <functional>

using namespace std;

// typedef
typedef long long LL;
typedef unsigned long long ULL;

// 
#define read freopen("in.txt","r",stdin)
#define write freopen("out.txt","w",stdout)

#define Z(a) (a<<1)
#define Y(a) (a>>1)

const double eps = 1e-6;
const double INFf = 1e100;
const int INFi = 1000000000;
const LL INFll = (LL)1<<62;
const double Pi = acos(-1.0);

template<class T> inline T sqr(T a){return a*a;}
template<class T> inline T TMAX(T x,T y)
{
	if(x>y) return x;
	return y;
}
template<class T> inline T TMIN(T x,T y)
{
	if(x<y) return x;
	return y;
}
template<class T> inline T MMAX(T x,T y,T z)
{
	return TMAX(TMAX(x,y),z);
}
template<class T> inline T MMIN(T x,T y,T z)
{
	return TMIN(TMIN(x,y),z);
}
template<class T> inline void SWAP(T &x,T &y)
{
	T t = x;
	x = y;
	y = t;
}


// code begin
int U[30011];

struct node
{
	int v;
	int dx;
	friend bool operator<(node a,node b)
	{
		return a.v<b.v;
	}
};
node data[30011];
int dx[30011];
int rdx[30011];
int Nx;
int M,N;
int ans[30011];
int Bit[30011];

inline int lowBit(int x)
{
	return x&(-x);
}

void update(int x,int c)
{
	for(int i=x;i<Nx;i+=lowBit(i))
	{
		Bit[i] += c;
	}
}

int getSum(int x)
{
	int sum = 0;
	for(int i=x;i>0;i-=lowBit(i))
	{
		sum += Bit[i];
	}
	return sum;
}

int main()
{
	read;
	write;
	while(scanf("%d %d",&M,&N)==2)
	{
		for(int i=1;i<=M;i++) 
		{
			scanf("%d",&data[i].v);
			data[i].dx = i;
		}
		for(int i=1;i<=N;i++) scanf("%d",U+i);
		
		//
		sort(data+1,data+M+1);
		Nx = 1;
		for(int i=1;i<=M;i++)
		{
			dx[data[i].dx] = Nx;
			rdx[Nx++] = data[i].v;
		}

		//
		memset(Bit,0,sizeof(Bit));
		int i=1,j=1,k=0,l,r,mid,tmp;
		for(;i<=M;i++)
		{
			// Add
			//printf("");
			update(dx[i],1);
			while(j<=N&&i==U[j])
			{
				//Get
				k++;
				// 找第k大
				l = 1;
				r = Nx;
				while(l<r)
				{
					mid = Y(l+r);
					tmp = getSum(mid);
					if(tmp>=k)
					{
						r = mid;
					}
					else
						l = mid+1;
				}
				ans[j] = rdx[r];
				j++;
			}
			if(j>N) break;
		}
		for(int i=1;i<=N;i++)
			printf("%d\n",ans[i]);
	}
	return 0;	
}