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<bits/stdc++.h>
using namespace std;
#define N 200020
typedef long long ll;
inline ll read(){
	ll x=0,f=1;
	char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9'){
		x=(x<<1)+(x<<3)+c-'0';
		c=getchar();
	}
	return x*f;
}
set<int> s;
int n,Q,p[N],f[N],dep[N],top[N],son[N],siz[N],cnt[N],vis[N];
#define int ll
ll dis[N],w[N],ans,a[N],b[N];
vector<int> G[N];
void dfs1(int u){
	siz[u]=1;
	for(auto v:G[u]){
		dep[v]=dep[u]+1;
		dfs1(v);
		siz[u]+=siz[v];
		if(!son[u]||siz[v]>siz[son[u]])son[u]=v;	
	}
}
void dfs2(int u,int topf){
	top[u]=topf;
	if(!son[u])return;
	dfs2(son[u],topf);
	for(auto v:G[u]){
		if(v^son[u])dfs2(v,v);
	}
}
int LCA(int x,int y){
	while(top[x]^top[y]){
		if(dep[top[x]]<dep[top[y]])swap(x,y);
		x=f[top[x]];
	}
	if(dep[x]>dep[y])swap(x,y);
	return x;
}
inline ll Dis(int x,int y){
	return dis[x]+dis[y]-2*dis[LCA(x,y)];
}
void Insert(int x){
	auto it=s.insert(x).first;
	int l=(*--it);
	++ ++it;
	int r=(it==s.end()?0:(*it));
	ans+=Dis(l,x)+Dis(x,r)-Dis(l,r);
}
void Del(int x){
	auto it=s.find(x);
	int l=(*--it);
	++ ++it;
	int r=(it==s.end()?0:(*it));
	ans-=Dis(l,x)+Dis(x,r)-Dis(l,r);
	s.erase(x);
}
signed main(){
	n=read(),Q=read();
	for(int i=1;i<=n;++i){
		p[i]=read();
	}
	for(int i=1;i<=n;++i){
		static int st[N],top;
		while(top&&p[i]>p[st[top]])--top;
		f[i]=st[top];
		G[f[i]].push_back(i);
		st[++top]=i;
	}
	for(int i=1;i<=n;++i)a[i]=read();
	for(int i=1;i<=n;++i)b[i]=read();
	for(int i=1;i<=n;++i){
		w[i]+=a[i]-b[i];
		w[f[i]]+=b[i];
	}
	dfs1(0);
	dfs2(0,0);
	for(int i=n;i>=1;--i){
		dis[i]+=w[i];
		dis[f[i]]+=min(0LL,dis[i]);
		dis[i]-=min(0LL,dis[i]);
	}
	dis[0]+=w[0];
	for(int i=1;i<=n;++i){
		dis[i]+=dis[f[i]];
	}
	++cnt[0];
	s.insert(0);
	while(Q--){
		int x=read();
		if(vis[x]){
			vis[x]=0;
			if(!--cnt[f[x]])Del(f[x]);
		}
		else{
			vis[x]=1;
			if(!cnt[f[x]]++)Insert(f[x]);
		}
		printf("%lld\n",(ans/2)+dis[0]);
	}
	return 0;
}
