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;

#define int long long

const int N=2e5+1e3+7;

int n,q,p[N],a[N],b[N],f[N],d[N];

int dep[N],anc[N][21],vis[N],cnt[N];

vector<int>g[N];

void dfs(int x)
{
    f[x]=a[x]-b[x];
    for(auto v:g[x])
    {
        dfs(v);
        f[x]+=b[v];
        if(f[v]<0)
            f[x]+=f[v],f[v]=0;
    }
}

int lca(int x,int y)
{
    if(dep[x]<dep[y])
        swap(x,y);
    for(int i=20;i>=0;i--)
        if(dep[x]-(1<<i)>=dep[y])
            x=anc[x][i];
    if(x==y)
        return x;
    for(int i=20;i>=0;i--)
        if(anc[x][i]!=anc[y][i])
            x=anc[x][i],y=anc[y][i];
    return anc[x][0];
}

int dis(int x,int y)
{
    return d[x]+d[y]-2*d[lca(x,y)];
}

set<int>s;

int ans;

void ins(int x)
{
    s.insert(x);
    auto it=s.find(x);
    int u=it==s.begin()?*--s.end():*prev(it);
    int v=it==--s.end()?*s.begin():*next(it);
    ans+=dis(u,x)+dis(x,v)-dis(u,v);
}

void del(int x)
{
    auto it=s.find(x);
    int u=it==s.begin()?*--s.end():*prev(it);
    int v=it==--s.end()?*s.begin():*next(it);
    ans-=dis(u,x)+dis(x,v)-dis(u,v);
    s.erase(it);
}

signed main()
{
    scanf("%lld%lld",&n,&q);
    for(int i=1;i<=n;i++)
        scanf("%lld",&p[i]);
    for(int i=1;i<=n;i++)
        scanf("%lld",&a[i]);
    for(int i=1;i<=n;i++)
        scanf("%lld",&b[i]);
    stack<int>st;
    p[0]=n+1;
    st.push(0);
    for(int i=1;i<=n;i++)
    {
        while(p[st.top()]<p[i])
            st.pop();
        g[st.top()].push_back(i);
        anc[i][0]=st.top();
        st.push(i);
    }
    dfs(0);
    for(int i=0;i<=n;i++)
        d[i]=d[anc[i][0]]+f[i],dep[i]=dep[anc[i][0]]+1;
    for(int j=1;j<=20;j++)
        for(int i=1;i<=n;i++)
            anc[i][j]=anc[anc[i][j-1]][j-1];
    ins(0);
    cnt[0]++;
    while(q--)
    {
        int x;
        scanf("%lld",&x);
        int y=anc[x][0];
        if(vis[x])
        {
            cnt[y]--;
            if(!cnt[y])
                del(y);
            vis[x]=0;
        }
        else
        {
            if(!cnt[y])
                ins(y);
            cnt[y]++;
            vis[x]=1;
        }
        printf("%lld\n",ans/2+f[0]);
    }
}