来源poj1442

Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions:

ADD (x): put element x into Black Box;

GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending.

Let us examine a possible sequence of 11 transactions:

Example 1

N Transaction i Black Box contents after transaction Answer

  (elements are arranged by non-descending)

1 ADD(3) 0 3

2 GET 1 3 3

3 ADD(1) 1 1, 3

4 GET 2 1, 3 3

5 ADD(-4) 2 -4, 1, 3

6 ADD(2) 2 -4, 1, 2, 3

7 ADD(8) 2 -4, 1, 2, 3, 8

8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8

9 GET 3 -1000, -4, 1, 2, 3, 8 1

10 GET 4 -1000, -4, 1, 2, 3, 8 2

11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8

It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type.

Let us describe the sequence of transactions by two integer arrays:

1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2).

2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6).

The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence.

Input

Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.

Output

Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.

Sample Input

7 4

3 1 -4 2 8 -1000 2

1 2 6 6

Sample Output

3

3

1

2

按上面一行输入，然后输入下面的的次数之后，输出第i个，i是从1开始，输出一次就加1；用两个优先队列，一个从小到大v2，一个从大到小v1，如果输入的数，比v1.top()大就推入2，或者空也推入，否则推入v1，然后把v1.top推入v2；

#include<iostream>

#include<stdio.h>

#include<stdlib.h>

#include <iomanip>

#include<cmath>

#include<float.h>

#include<string.h>

#include<algorithm>

#define sf scanf

#define pf printf

#define scf(x) scanf("%d",&x)

#define scff(x,y) scanf("%d%d",&x,&y)

#define prf(x) printf("%d\n",x)

#define mm(x,b) memset((x),(b),sizeof(x))

#include<vector>

#include<queue>

#include<map>

#define rep(i,a,n) for (int i=a;i<n;i++)

#define per(i,a,n) for (int i=a;i>=n;i--)

typedef long long ll;

const ll mod=1e9+7;

const double eps=1e-8;

const int inf=0x3f3f3f3f;

using namespace std;

const double pi=acos(-1.0);

const int N=3e4+5;

priority_queue <ll> v1;//从大到小排

priority_queue <ll,vector<ll>,greater<ll> > v2;

ll num[N];

int main()

{

	ll n,m,c=1;

	ll ans;

	sf("%lld%lld",&n,&m);

	rep(i,0,n)

	sf("%lld",&num[i]);

	int i=0;

	while(m--)

	{

		scf(c);

		for(;i<c;i++)

		{

			if(v1.empty()||v1.top()<num[i])

			v2.push(num[i]);

			else

			{

				v1.push(num[i]);

				int temp=v1.top();

				v2.push(temp);

				v1.pop();

			}

		}

		int ans=v2.top();

		v2.pop();

		v1.push(ans);

		prf(ans);

	}

	return 0;

}