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.
// Author -- xyr2005

#include<bits/stdc++.h>

#define lowbit(x) ((x)&(-(x)))
#define DEBUG fprintf(stderr,"Running on Line %d in Function %s\n",__LINE__,__FUNCTION__)
#define SZ(x) ((int)x.size())
#define mkpr std::make_pair
#define pb push_back

typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef std::pair<int,int> pi;
typedef std::pair<ll,ll> pl;
using std::min;
using std::max;

const int inf=0x3f3f3f3f,Inf=0x7fffffff;
const ll INF=0x3f3f3f3f3f3f3f3f;

std::mt19937 rnd(std::chrono::steady_clock::now().time_since_epoch().count());
template <typename _Tp>_Tp gcd(const _Tp &a,const _Tp &b){return !b?a:gcd(b,a%b);}
template <typename _Tp>inline _Tp abs(const _Tp &a){return a>=0?a:-a;}
template <typename _Tp>inline void chmax(_Tp &a,const _Tp &b){a=a<b?b:a;}
template <typename _Tp>inline void chmin(_Tp &a,const _Tp &b){a=a<b?a:b;}
template <typename _Tp>inline void read(_Tp &x)
{
	char ch(getchar());bool f(false);while(!isdigit(ch)) f|=ch==45,ch=getchar();
	x=ch&15,ch=getchar();while(isdigit(ch)) x=(((x<<2)+x)<<1)+(ch&15),ch=getchar();
	f&&(x=-x);
}
template <typename _Tp,typename... Args>void read(_Tp &t,Args &...args){read(t);read(args...);}
inline int read_str(char *s)
{
	char ch(getchar());while(ch==' '||ch=='\r'||ch=='\n') ch=getchar();
	char *tar=s;*tar++=ch,ch=getchar();while(ch!=' '&&ch!='\r'&&ch!='\n'&&ch!=EOF) *tar++=ch,ch=getchar();
	return *tar=0,tar-s;
}

const int N=200005;
int p[N],a[N],b[N],fa[N],st[N],top,anc[N][20],dep[N],cnt[N];
bool vis[N];
ll dis[N];
int lca(int x,int y)
{
	if(dep[x]<dep[y])std::swap(x,y);
	for(int i=19;i>=0;--i)if(dep[anc[x][i]]>=dep[y])x=anc[x][i];
	if(x==y)return x;
	for(int i=19;i>=0;--i)if(anc[x][i]!=anc[y][i])x=anc[x][i],y=anc[y][i];
	return fa[x];
}
inline ll getdis(int x,int y){return dis[x]+dis[y]-2LL*dis[lca(x,y)];}
std::set<int> S;
ll ans;
void ins(int x)
{
	auto it=S.insert(x).first,qwq(it);
	int pre=it==S.begin()?*S.rbegin():*--it;
	int suf=++qwq==S.end()?*S.begin():*qwq;
	ans-=getdis(pre,suf),ans+=getdis(x,pre),ans+=getdis(x,suf);
}
void del(int x)
{
	auto it=S.find(x),qwq(it);
	int pre=it==S.begin()?*S.rbegin():*--it;
	int suf=++qwq==S.end()?*S.begin():*qwq;
	ans+=getdis(pre,suf),ans-=getdis(x,pre),ans-=getdis(x,suf),S.erase(x);
}
int main()
{
	int n,q;read(n,q);
	for(int i=1;i<=n;++i)read(p[i]);
	for(int i=1;i<=n;++i)
	{
		while(top&&p[i]>p[st[top]])--top;
		fa[i]=st[top],st[++top]=i,anc[i][0]=fa[i],dep[i]=dep[fa[i]]+1;
		for(int j=1;j<20;++j)anc[i][j]=anc[anc[i][j-1]][j-1];
	}
	for(int i=1;i<=n;++i)read(a[i]),dis[i]=a[i];
	for(int i=1;i<=n;++i)read(b[i]),dis[i]-=b[i],dis[fa[i]]+=b[i];
	for(int i=n;i>=1;--i)dis[fa[i]]+=min(0LL,dis[i]),chmax(dis[i],0LL);
	for(int i=1;i<=n;++i)dis[i]+=dis[fa[i]];
	int x;S={0},cnt[0]=inf;
	while(q--)
	{
		read(x);
		if(vis[x])!--cnt[fa[x]]?del(fa[x]):void();
		else !cnt[fa[x]]++?ins(fa[x]):void();
		vis[x]^=1;
		printf("%lld\n",ans/2+dis[0]);
	}
	return 0;
}
