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<set>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ch() getchar()
#define pc(x) putchar(x)
using namespace std;
template<typename T>void read(T&x){
	static char c;static int f;
	for(c=ch(),f=1;c<'0'||c>'9';c=ch())if(c=='-')f=-f;
	for(x=0;c>='0'&&c<='9';c=ch())x=x*10+(c&15);x*=f;
}
template<typename T>void write(T x){
	static char q[65];int cnt=0;
	if(x<0)pc('-'),x=-x;
	q[++cnt]=x%10,x/=10;
	while(x)
		q[++cnt]=x%10,x/=10;
	while(cnt)pc(q[cnt--]+'0');
}
const int maxn=200005;
int p[maxn],st[maxn],tp,pa[maxn][22],dp[maxn];
long long a[maxn],b[maxn],val[maxn],dis[maxn];
int lca(int x,int y){
	if(dp[x]<dp[y])x^=y^=x^=y;
	for(int t=dp[x]-dp[y],cn=0;t;t>>=1,++cn)
		if(t&1)x=pa[x][cn];
	if(x==y)return x;
	for(int t=20;t>=0;--t)
		if(pa[x][t]!=pa[y][t])
			x=pa[x][t],y=pa[y][t];
	return pa[x][0];
}
int vis[maxn],cnt[maxn];set<int>s;
long long sum;
long long dist(int x,int y){
	return dis[x]+dis[y]-2*dis[lca(x,y)];
}
set<int>::iterator it;
void ins(int x){
	it=s.insert(x).first;int l=*(--it);++it;++it;
	int r=((it==s.end())?*s.begin():*it);
	sum+=dist(l,x)+dist(x,r)-dist(l,r);
}
void del(int x){
	it=s.find(x);int l=*(--it);++it;++it;
	int r=((it==s.end())?*s.begin():*it);
	sum-=dist(l,x)+dist(x,r)-dist(l,r);
	--it;s.erase(it);
}
int main(){
	int n,q;read(n),read(q);
	for(int i=1;i<=n;++i){
		read(p[i]);
		while(tp&&p[st[tp]]<p[i])--tp;
		dp[i]=dp[pa[i][0]=st[tp]]+1;st[++tp]=i;
		for(int j=1;(1<<j)<=dp[i];++j)
			pa[i][j]=pa[pa[i][j-1]][j-1];
	}
	for(int i=1;i<=n;++i)read(a[i]),val[i]+=a[i];
	for(int i=1;i<=n;++i)read(b[i]),val[i]-=b[i],val[pa[i][0]]+=b[i];
	for(int i=n;i>=1;--i)dis[pa[i][0]]+=min(0ll,dis[i]+=val[i]),dis[i]-=min(0ll,dis[i]);
	dis[0]+=val[0];for(int i=1;i<=n;++i)dis[i]+=dis[pa[i][0]];cnt[0]=1;s.insert(0);
	while(q--){
		int x;read(x);
		if(vis[x]){
			vis[x]=false;
			if(!(--cnt[pa[x][0]]))
				del(pa[x][0]);
		}
		else{
			vis[x]=true;
			if(!(cnt[pa[x][0]]++))
				ins(pa[x][0]);
		}
		write(sum/2+dis[0]),pc('\n');
	}
	return 0;
}