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>
#define For(i,a,b) for(register int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(register int i=(a);i>=(b);--i)
#define int long long
using namespace std;
inline int read()
{
    char c=getchar();int x=0;bool f=0;
    for(;!isdigit(c);c=getchar())f^=!(c^45);
    for(;isdigit(c);c=getchar())x=(x<<1)+(x<<3)+(c^48);
    if(f)x=-x;return x;
}

#define maxn 200005
int n,m;
int p[maxn],stk[maxn],top;
int fa[maxn][21],dis[maxn],dep[maxn];
int a[maxn],b[maxn],c[maxn]; 
int cnt[maxn];

inline int lca(int u,int v){
	if(dep[u]<dep[v])swap(u,v);
	int gap=dep[u]-dep[v];
	Rep(i,20,0)if(gap>>i&1)u=fa[u][i];
	if(u==v)return u;
	Rep(i,20,0)if(fa[u][i]!=fa[v][i])u=fa[u][i],v=fa[v][i];
	return fa[u][0];
}
inline int dist(int u,int v){
	return dis[u]+dis[v]-(dis[lca(u,v)]<<1);
}

int dp[maxn],val[maxn];
set<int>s;
bool in[maxn];
int now;
void ins(int x){
	auto it=s.insert(x).first;int l=*(--it);++it;++it;
	int r=((it==s.end())?*s.begin():*it);
	now+=dist(l,x)+dist(x,r)-dist(l,r);
}
void del(int x){
	auto it=s.find(x);int l=*(--it);++it;++it;
	int r=((it==s.end())?*s.begin():*it);
	now-=dist(l,x)+dist(x,r)-dist(l,r);
	--it;s.erase(it);
}

signed main()
{
	n=read(),m=read();
	For(i,1,n){
		p[i]=read();
		while(top&&p[stk[top]]<p[i])top--;
		fa[i][0]=stk[top];
		dep[i]=dep[fa[i][0]]+1;
		For(j,1,20)fa[i][j]=fa[fa[i][j-1]][j-1];
		stk[++top]=i;
	}
	// 这样建树后
	// 一个点要经过 == 父亲选A
	For(i,1,n)a[i]=read(),c[i]=a[i];
	For(i,1,n)b[i]=read(),c[i]-=b[i],c[fa[i][0]]+=b[i];
	Rep(i,n,1){
//		dp[i]+=c[i];
//		if(i)dp[fa[i][0]]+=min(dp[i],0ll);
		dis[i]+=c[i];
		dis[fa[i][0]]+=min(0ll,dis[i]);
		dis[i]=max(dis[i],0ll);
	}
	
	dis[0]+=c[0];
//	For(i,0,n)cout<<dis[i]<<' ';puts("");
	For(i,1,n)dis[i]+=dis[fa[i][0]]; 
	
//	val[0]=dp[0];
//	For(i,1,n)val[i]=max(0ll,dp[i]);
	
//	For(i,1,n)dis[i]=dis[fa[i][0]]+val[i];
//	For(i,1,n)cout<<dp[i]<<' ';puts("");For(i,1,n)cout<<val[i]<<" ";puts("");
	cnt[0]=1,s.insert(0);
	For(_,1,m)
	{
		int u=read(),f=fa[u][0];
		if(in[u]){
			in[u]=0;
			cnt[f]--;
			if(!cnt[f])del(f);
		}else{
			in[u]=1;
			cnt[f]++;
			if(cnt[f]==1)ins(f);
		}
		cout<<now/2+dis[0]<<endl;
	}
    return 0;
}