Problem: There are n squares drawn from left to right on the floor. The i-th square has three integers p_i,a_i,b_i, written on it. The sequence p_1,p_2,...,p_n forms a permutation.

Each round you will start from the leftmost square 1 and jump to the right. If you are now on the i-th square, you can do one of the following two operations:

  1. Jump to the i+1-th square and pay the cost a_i. If i=n, then you can end the round and pay the cost a_i. 
  2. Jump to the j-th square and pay the cost b_i, where j is the leftmost square that satisfies j > i, p_j > p_i. If there is no such j then you can end the round and pay the cost b_i. 



There are q rounds in the game. To make the game more difficult, you need to maintain a square set S (initially it is empty). You must pass through these squares during the round (other squares can also be passed through). The square set S for the i-th round is obtained by adding or removing a square from the square set for the (i-1)-th round. 

For each round find the minimum cost you should pay to end it.

Input

The first line contains two integers n, q (1≤ n,q≤ 2 ⋅ 10^5) — the number of squares and the number of rounds.

The second line contains n distinct integers p_1,p_2,...,p_n (1≤ p_i≤ n). It is guaranteed that the sequence p_1,p_2,...,p_n forms a permutation.

The third line contains n integers a_1,a_2,…,a_n (-10^9≤ a_i≤ 10^9).

The fourth line contains n integers b_1,b_2,...,b_n (-10^9≤ b_i≤ 10^9).

Then q lines follow, i-th of them contains a single integer x_i (1≤ x_i≤ n). If x_i was in the set S on the (i-1)-th round you should remove it, otherwise, you should add it.

Output

Print q lines, each of them should contain a single integer — the minimum cost you should pay to end the corresponding round.

Examples

Input


3 2
2 1 3
10 -5 4
3 -2 3
1
2


Output


6
8


Input


5 4
2 1 5 3 4
6 -5 3 -10 -1
0 3 2 7 2
1
2
3
2


Output


-8
-7
-7
-8

Note

Let's consider the character T as the end of a round. Then we can draw two graphs for the first and the second test.

<image> <image>

In the first round of the first test, the set that you must pass through is \{1\}. The path you can use is 1→ 3→ T and its cost is 6.

In the second round of the first test, the set that you must pass through is \{1,2\}. The path you can use is 1→ 2→ 3→ T and its cost is 8.

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
long long a,b,c,d,e,i,j,ii,jj,zx,xc,tes,t,f[200009],A[200009],B[200009],rg[200009][20],rgx[200009][20],lf[200009][20],lfx[200009][20],P[200009],pas;
stack <long long> st;
set <long long> s;
set <long long>::iterator it,lft,rgt;
long long uprg(long long q, long long w){
	if(q>=w) return 0;
	long long jm=0;
	long long h=0;
	for(h=18; h>=0; h--){
		if(rg[q][h]!=0&&rg[q][h]<w){
			jm+=rgx[q][h];
			q=rg[q][h];
		}
	}
	jm+=rgx[q][0];
	return jm;
}
long long uplf(long long q, long long w){
	if(w<=q) return 0;
	long long jm=0;
	long long h=0;
	for(h=19; h>=0; h--){
		if(lf[w][h]!=0&&lf[w][h]>q){
			jm+=lfx[w][h];
			w=lf[w][h];
		}
	}
	jm+=lfx[w][0];
	return jm;
}
long long fun(long long q, long long w){
	if(q>=w) return 0;
	long long jm=0;
	long long h=0;
	for(h=18; h>=0; h--){
		if(rg[q][h]!=0&&rg[q][h]<=w){
			jm+=rgx[q][h];
			q=rg[q][h];
		}
	}
	if(q==w) return jm;
	jm+=uplf(q,w);
	return jm;
}
int main(){
	ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	cin>>a>>tes;
	for(i=1; i<=a; i++) cin>>f[i];
	for(i=1; i<=a; i++) cin>>A[i];
	for(i=1; i<=a; i++) cin>>B[i];
	while(st.size()) st.pop();
	st.push(a+1);
	f[a+1]=a+2;
	for(i=a; i>=1; i--){
		while(st.size()>0&&f[st.top()]<f[i]) st.pop();
		if(st.size()==0){
			
		}else{
			rg[i][0]=st.top();
			rgx[i][0]=min(B[i],A[i]+uprg(i+1,st.top()));
		}
		for(j=1; j<=18; j++){
			rg[i][j]=rg[rg[i][j-1]][j-1];
			rgx[i][j]=rgx[i][j-1]+rgx[rg[i][j-1]][j-1];
		}
		st.push(i);
	}
	while(st.size()) st.pop();
	for(i=1; i<=a; i++){
		while(st.size()>0&&f[st.top()]<f[i]) st.pop();
		if(st.size()==0){
			
		}else{
			lf[i][0]=st.top();
			lfx[i][0]=A[st.top()]+uprg(st.top()+1,i);
		}
		for(j=1; j<=18; j++){
			lf[i][j]=lf[lf[i][j-1]][j-1];
			lfx[i][j]=lfx[i][j-1]+lfx[lf[i][j-1]][j-1];
		}
		st.push(i);
	}
	s.insert(1);
	s.insert(a+1);
	P[1]=fun(1,a+1);
	pas+=P[1];
	for(t=1; t<=tes; t++){
		cin>>c;
		if(c==1){
			cout<<pas<<endl;
			continue;
		}
		it=s.lower_bound(c);
		if(it==s.end()||(*it)!=c){
			P[c]=fun(c,(*it));
			pas+=P[c];
			lft=it;lft--;
			pas-=P[(*lft)];
			P[(*lft)]=fun((*lft),c);
			pas+=P[(*lft)];
			s.insert(c);
		}else{
			lft=it;lft--;
			pas-=P[(*lft)];
			rgt=it;rgt++;
			pas-=P[c];
			P[(*lft)]=fun((*lft),(*rgt));
			pas+=P[(*lft)];
			s.erase(it);
		}
		cout<<pas<<endl;
	}
	P[0]=2*6;
	pas=P[0]+5;
	return 0;
}