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.
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string.h>
#include<vector>
#include<queue>
#include<map>
#include<ctime>
#include<set>
#define fi first
#define se second
#define mp make_pair
#define pi pair<int,int>
#define pb push_back
#define pl pair<int,long long>
#define vi vector<int>
#define vl vector<long long>
#define int long long
using namespace std;
typedef long long ll;
const int N=4e5+55;
int n,m,a[N],b[N],p[N],val[N],st[N],head,vis[N],x;
vi v[N];
int deep[N],son[N],siz[N],top[N],fu[N],dis[N];
void link(int x,int y){
	v[x].pb(y);
}
void dfs(int u,int fa){
	siz[u]=1;deep[u]=deep[fa]+1;
	for(unsigned int i=0;i<v[u].size();i++){
		if(v[u][i]==fa)continue;
		dfs(v[u][i],u);
		siz[u]+=siz[v[u][i]];
		if(dis[v[u][i]]<0)dis[u]+=dis[v[u][i]],dis[v[u][i]]=0;
		if(siz[v[u][i]]>siz[son[u]])son[u]=v[u][i];
	}
	dis[u]+=val[u];
}
void dfs1(int u,int fa){
	top[u]=fa;
	if(u)dis[u]+=dis[fu[u]];
	if(son[u])dfs1(son[u],fa);
	for(unsigned int i=0;i<v[u].size();i++){
		if(top[v[u][i]])continue;
		dfs1(v[u][i],v[u][i]);
	}
}
int lca(int x,int y){
	while(top[x]!=top[y]){
		if(deep[top[x]]<deep[top[y]])swap(x,y);
		x=fu[top[x]];
	}
	if(deep[x]<deep[y])swap(x,y);
	return y;
}
int Dis(int x,int y){
	return dis[x]+dis[y]-2ll*dis[lca(x,y)];
}
set<int>s;
set<int>::iterator it,iv;
int da=0;
int fq[N];
void add(int x){
	s.insert(x);
	it=s.lower_bound(x);
	int L=*(--it);
	it++;it++;
	int R;
	if(it==s.end())it=s.begin(),R=*it;
	else R=*it;
	da+=(Dis(L,x)+Dis(R,x)-Dis(L,R));
}
void del(int x){
	iv=it=s.lower_bound(x);
	it--;
	//s.erase(iv);
	int L=*(it);
	it++;it++;
	int R;
	if(it==s.end())it=s.begin(),R=*it;
	else R=*it;
	da-=(Dis(L,x)+Dis(R,x)-Dis(L,R));
	s.erase(iv);
}
int zero[N];
signed main(){
	scanf("%lld%lld",&n,&m);
	for(int i=1;i<=n;i++)scanf("%lld",&p[i]);
	for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
	for(int i=1;i<=n;i++)scanf("%lld",&b[i]);
	p[0]=1e9+7;st[++head]=0;
	for(int i=1;i<=n;i++){
		while(head and p[st[head]]<p[i])head--;
		link(st[head],i);fu[i]=st[head];
		st[++head]=i;
	}
	for(int i=1;i<=n;i++){
		val[i]=a[i]-b[i];
		val[fu[i]]+=b[i];
	}
	dfs(0,0);
	dfs1(0,0);
	int ans1=dis[0];
	s.insert(0);
	fq[0]=1;
	for(int i=1;i<=m;i++){
		scanf("%lld",&x);
		if(!vis[x]){
			fq[fu[x]]++;
			if(fq[fu[x]]==1){
				add(fu[x]);
				
			}vis[x]=1;
		}
		else{
			fq[fu[x]]--;
			if(!fq[fu[x]]){
				del(fu[x]);
				
			}vis[x]=0;
		}
		printf("%lld\n",ans1+da/2ll);
	}
	return 0;
}
