Problem: 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.

Solution (in CPP):

Solution:
#include<bits/stdc++.h>

#define y1 dmytxdy
#define pb push_back
#define fi first
#define se second
#define mp make_pair

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef long double ld;

template <typename T> bool chkmin(T &x,T y){return x>y?x=y,1:0;}
template <typename T> bool chkmax(T &x,T y){return x<y?x=y,1:0;}

int readint(){
	int x=0,f=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}

int n,q,top;
ll p[200005],a[200005],b[200005],s[200005],fa[200005],ch[200005][2],rev[200005],sum[200005],val[200005],stk[200005],ans[200005];
vector<pii> qry[200005];

bool nroot(int x){return ch[fa[x]][0]==x||ch[fa[x]][1]==x;}
bool son(int x){return ch[fa[x]][1]==x;}
void reverse(int x){rev[x]^=1,swap(ch[x][0],ch[x][1]);}
void pushdown(int x){if(rev[x]) reverse(ch[x][0]),reverse(ch[x][1]),rev[x]=0;}
void update(int x){sum[x]=val[x]+sum[ch[x][0]]+sum[ch[x][1]];}

void rotate(int x){
	int y=fa[x],z=fa[y],k=son(x),w=ch[x][!k];
	if(nroot(y)) ch[z][son(y)]=x;
	ch[x][!k]=y,ch[y][k]=w;
	if(w) fa[w]=y;
	fa[y]=x,fa[x]=z;
	update(y),update(x);
}

void pushall(int x){
	if(nroot(x)) pushall(fa[x]);
	pushdown(x);
}

void splay(int x){
	pushall(x);
	while(nroot(x)){
		int y=fa[x];
		if(nroot(y)) rotate(son(x)==son(y)?y:x);
		rotate(x);
	}
}

void access(int x){for(int y=0;x;x=fa[y=x]) splay(x),ch[x][1]=y,update(x);}
void makeroot(int x){access(x); splay(x); reverse(x);}
void cut(int x,int y){makeroot(x); access(y); splay(x); assert(ch[x][1]==y); ch[x][1]=0,fa[y]=0; update(x);}

int main(){
	n=readint(); q=readint();
	for(int i=1;i<=n;i++) p[i]=readint();
	for(int i=1;i<=n;i++) a[i]=readint(),s[i]=s[i-1]+a[i];
	for(int i=1;i<=n;i++) b[i]=readint();
	set<int> st;
	st.insert(1),st.insert(n+1);
	qry[n+1].pb(mp(1,1));
	int x;
	for(int i=1;i<=q;i++){
		x=readint();
		if(x==1) continue;
		if(st.find(x)!=st.end()){
			st.erase(x);
			auto it2=st.lower_bound(x),it1=it2; it1--;
			qry[x].pb(mp(*it1,-i));
			qry[*it2].pb(mp(x,-i));
			qry[*it2].pb(mp(*it1,i));
		}
		else{
			auto it2=st.lower_bound(x),it1=it2; it1--;
			qry[x].pb(mp(*it1,i));
			qry[*it2].pb(mp(x,i));
			qry[*it2].pb(mp(*it1,-i));
			st.insert(x);
		}
	}
	top=0;
	p[n+1]=n+1;
	for(int i=1;i<=n+1;i++){
		if(top&&p[i-1]>p[i]){
			makeroot(i-1);
			val[i-1]=a[i-1];
			update(i-1);
			fa[i-1]=i;
			makeroot(i);
		}
		while(top&&p[stk[top]]<p[i]){
			if(stk[top]==i-1){
				makeroot(i-1);
				val[i-1]=min(a[i-1],b[i-1]);
				update(i-1);
				fa[i-1]=i;
				makeroot(i);
			}
			else{
				access(stk[top]); splay(stk[top]);
				if(sum[stk[top]]>b[stk[top]]){
					cut(stk[top],stk[top]+1);
					val[stk[top]]=b[stk[top]];
					update(stk[top]);
					fa[stk[top]]=i;
					makeroot(i);
				}
			}
			top--;
		}
		stk[++top]=i;
		for(auto j:qry[i]){
			access(j.fi); splay(j.fi);
			if(j.se>0) ans[j.se]+=sum[j.fi];
			else ans[-j.se]-=sum[j.fi];
		}
	}
	for(int i=1;i<=q;i++) ans[i]+=ans[i-1];
	for(int i=1;i<=q;i++) printf("%lld\n",ans[i]);
	return 0;
}