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;
const int N=2e5+5;
typedef long long ll;
typedef pair<int,int> pii;
#define fi first
#define se second
inline int read(){
	int s=0,f=1;char ch=getchar();
	while(!isdigit(ch)){if(ch=='-') f=-1;ch=getchar();}
	while(isdigit(ch)) s=s*10+ch-'0',ch=getchar();
	return s*f;
}
int n,m;
int p[N],to[N];ll a[N],b[N];
ll ans[N];
int top;pii stk[N];
set<int> s;
vector<pii> q[N];
inline void add(int l,int r,int id){
	q[l].emplace_back(r,id);
}
struct Bit{
	ll c[N];
	inline void add(int x,ll y){
		for(;x<=n+1;x+=x&-x) c[x]+=y;
	}
	inline ll query(int x){
		ll res=0;
		for(;x;x-=x&-x) res+=c[x];
		return res;
	}
}tr;
signed main(){
	n=read(),m=read();
	for(int i=1;i<=n;i++) p[i]=read();
	for(int i=1;i<=n;i++) a[i]=read()+a[i-1];
	for(int i=1;i<=n;i++) b[i]=read();
	stk[++top]={n+1,n+1};
	for(int i=n;i>=1;i--){
		int v=upper_bound(stk+1,stk+top+1,pii(p[i],-1),greater<pii>())-stk-1;
		to[i]=stk[v].se;//cout<<to[i]<<endl;
		b[i]=b[i]-(a[to[i]-1]-a[i-1]);
		while(top&&p[i]>stk[top].fi) top--;
		stk[++top]={p[i],i};
	}
	s.insert(1);s.insert(n+1);
	add(1,n+1,0);
	for(int i=1;i<=m;i++){
		int x=read();
		if(x==1) continue;
		if(!s.count(x)){
			auto it=s.lower_bound(x);
			int l=*prev(it),r=*it;
			add(l,r,-i);add(l,x,i);add(x,r,i);
			s.insert(x);
		}else {
			s.erase(x);
			auto it=s.lower_bound(x);
			int l=*prev(it),r=*it;
			add(l,r,i);add(l,x,-i);add(x,r,-i); 
		}
	}
	for(int i=n;i>=1;i--){
		ll v=tr.query(to[i]);
		if(b[i]<v) tr.add(to[i],b[i]-v);
		for(auto qry:q[i]){
			ll t=tr.query(qry.fi);
			if(qry.se>=0) ans[qry.se]+=t+a[qry.fi-1]-a[i-1];
			else ans[-qry.se]-=t+a[qry.fi-1]-a[i-1];
		} 
	}
	for(int i=1;i<=m;i++) ans[i]+=ans[i-1],printf("%lld\n",ans[i]);
	return 0;
}