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.
//zxggtxdy!!!
#include<bits/stdc++.h>
using namespace std;
const int N=1e6+7; long long a[N],b[N],f[N],g[N],val[N],sum[N];
int n,q,k,p[N],G[N],P[N],nxt[N]; vector<int>E[N]; bool fl[N];
struct node{
	int x,y,f,g;
}e[N];
namespace LCT{
	bool rev[N]; int st[N],fa[N],ch[N][2];
	inline void upd(int u){
		g[u]=g[ch[u][0]]+g[ch[u][1]]+f[u]; return;
	}
	inline bool isroot(int u){
		return (ch[fa[u]][0]!=u&&ch[fa[u]][1]!=u);
	}
	inline bool get(int u){
		return (ch[fa[u]][1]==u);
	}
	inline void pushdown(int u){
		if(rev[u]) rev[ch[u][0]]^=1,rev[ch[u][1]]^=1,swap(ch[u][0],ch[u][1]),rev[u]=0;
	}
	inline void rotate(int u){
		int pa=fa[u],gr=fa[pa],x=get(u);
		if(!isroot(pa)) ch[gr][get(pa)]=u; fa[u]=gr;
		ch[pa][x]=ch[u][x^1],fa[ch[pa][x]]=pa,ch[u][x^1]=pa,fa[pa]=u,upd(pa),upd(u),upd(gr);
	}
	inline void splay(int u){
		st[st[0]=1]=u;
		for(int i=u;!isroot(i);i=fa[i]) st[++st[0]]=fa[i];
		for(int i=st[0];i>=1;i--) pushdown(st[i]);
		for(;!isroot(u);rotate(u)) if(!isroot(fa[u])) rotate(get(fa[u])==get(u)?fa[u]:u);
	}
	inline void access(int u){
		for(int i=0;u>0;i=u,u=fa[u]) splay(u),ch[u][1]=i,upd(u);
	}
	inline void makeroot(int u){
		access(u),splay(u),rev[u]^=1;
	}
	inline void link(int u,int t){
		makeroot(u),makeroot(t),fa[t]=u;
	}
	inline void cut(int u,int t){
		makeroot(u),access(t),splay(t),ch[t][0]=0,fa[u]=0,upd(t);
	}
	inline long long getans(int u,int t){
		makeroot(u),access(t),splay(t); return g[t];
	}
}
inline bool cmp(node A,node B){
	return A.y<B.y;
}
int main(){
	cin>>n>>q;
	for(int i=1;i<=n;i++) scanf("%d",&p[i]);
	for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
	for(int i=1;i<=n;i++) scanf("%lld",&b[i]);
	for(int i=n;i>=1;i--){
		if(!G[0]||p[G[1]]<=p[i]) nxt[i]=0;
		else{
			int l=1,r=G[0];
			while(l<=r){
				int d=(l+r)>>1;
				if(p[G[d]]>p[i]) nxt[i]=G[d],l=d+1; else r=d-1;
			}
		}
		while(G[0]&&p[G[G[0]]]<=p[i]) G[0]--; G[++G[0]]=i;
	}
	for(int i=1;i<=n;i++)
		if(nxt[i]&&nxt[i]==i+1) a[i]=min(a[i],b[i]);
		else E[nxt[i]].push_back(i);
	k=n;
	for(int i=n;i>=1;i--)
		val[i]=min(val[nxt[i]]+b[i],val[i+1]+a[i]);
	set<int>F; F.insert(1),F.insert(n+1); int m=0;
	for(int i=1;i<=q;i++){
		int x; scanf("%d",&x); if(x==1) continue;
		if(!fl[x]){
			F.insert(x);
			int A=(*--F.find(x)),B=(*++F.find(x));
			m++,e[m].x=A,e[m].y=B,e[m].f=i,e[m].g=-1;
			m++,e[m].x=A,e[m].y=x,e[m].f=i,e[m].g=+1;
			m++,e[m].x=x,e[m].y=B,e[m].f=i,e[m].g=+1;
		}
		else{
			int A=(*--F.find(x)),B=(*++F.find(x));
			m++,e[m].x=A,e[m].y=B,e[m].f=i,e[m].g=+1;
			m++,e[m].x=A,e[m].y=x,e[m].f=i,e[m].g=-1;
			m++,e[m].x=x,e[m].y=B,e[m].f=i,e[m].g=-1;
			F.erase(x);
		}
		fl[x]^=1;
	}
	sort(e+1,e+m+1,cmp); int j=1;
	for(int i=2;i<=n;i++){
		P[i-1]=++k,f[k]=g[k]=a[i-1],LCT::link(i,k),LCT::link(i-1,k);
		for(int j=(int)(E[i].size())-1;j>=0;j--){
			int x=E[i][j]; long long q=min(LCT::getans(x,i),b[x]);
				LCT::cut(x,P[x]),LCT::cut(x+1,P[x]);
				k++,f[k]=g[k]=q,LCT::link(i,k),LCT::link(x,k);
		}
		while(e[j].y==i)
			sum[e[j].f]+=e[j].g*LCT::getans(e[j].x,e[j].y),j++;
	}
	while(j<=m) sum[e[j].f]+=e[j].g*val[e[j].x],j++; sum[0]=val[1];
	for(int i=1;i<=q;i++) sum[i]+=sum[i-1],printf("%lld\n",sum[i]);
	return 0;
}