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;
typedef long long LL;
const int N = 222222;
int n,q,sz,p[N],a[N],b[N],c[N];
vector<int> v;
LL f[N],sa[N],sf[N],t[N];
set<int> S;
set<int>::iterator it;
int qx[N],ql[N],qr[N],qt[N];
vector<pair<int,int> > h[N];
LL ans[N];
void ad(int x,LL y){
	while(x<=n+1){
		t[x]+=y;
		x+=x&-x;
	}
}
LL qu(int x){
	LL y=0;
	while(x){
		y+=t[x];
		x-=x&-x;
	}
	return y;
}
int main(){
	int i,j,k,o;
	LL x,y;
	scanf("%d%d",&n,&q);
	for(i=1;i<=n;i++)
		scanf("%d",p+i);
	for(i=1;i<=n;i++)
		scanf("%d",a+i);
	for(i=1;i<=n;i++)
		scanf("%d",b+i);
	p[0]=n+1;
	p[n+1]=n+1;
	v.push_back(0);
	sz=0;
	for(i=1;i<=n+1;i++){
		while(p[v[sz]]<p[i]){
			c[v[sz]]=i;
			v.pop_back();
			sz--;
		}
		v.push_back(i);
		sz++;
	}
	for(i=n;i>=1;i--){
		sa[i]=sa[i+1]+a[i];
		x=sa[i]-sa[c[i]]-b[i];
		y=sf[i+1]-sf[c[i]];
		if(x>y)
			f[i]=x-y;
		else
			f[i]=0;
		sf[i]=sf[i+1]+f[i];
	}
	S.insert(0);
	S.insert(n+1);
	h[0].push_back(make_pair(n+1,0));
	for(i=1;i<=q;i++){
		scanf("%d",qx+i);
		if(S.find(qx[i])==S.end()){
			qt[i]=1;
			it=S.lower_bound(qx[i]);
			qr[i]=*it;
			it--;
			ql[i]=*it;
			S.insert(qx[i]);
		}
		else{
			qt[i]=-1;
			S.erase(qx[i]);
			it=S.lower_bound(qx[i]);
			qr[i]=*it;
			it--;
			ql[i]=*it;
		}//cout<<ql[i]<<qx[i]<<qr[i]<<endl;
		h[ql[i]].push_back(make_pair(qx[i],i));
		h[qx[i]].push_back(make_pair(qr[i],i));
		h[ql[i]].push_back(make_pair(qr[i],-i));
	}
	for(i=n;i>=0;i--){
		if(i>0)
			ad(c[i],f[i]);
		for(j=0;j<h[i].size();j++){
			k=h[i][j].first;
			o=h[i][j].second;
			if(o>=0)
				ans[o]+=qu(k);
			else
				ans[-o]-=qu(k);
		}
	}
	x=ans[0];
	for(i=1;i<=q;i++){
		x+=ans[i]*qt[i];
		cout<<sa[1]-x<<endl;
	}
	return 0;
}
