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 c=200002, k=20;
int n, q, p[c], szint[c], fel[c][k], f[c], db[c];
long long aa[c], bb[c], dist[c], val[c], sum;
bool v[c];
set<int> s;
int lca(int a, int b) {
    if (szint[a]<szint[b]) {
        swap(a, b);
    }
    for (int i=k-1; i>=0; i--) {
        if (szint[fel[a][i]]>=szint[b]) {
            a=fel[a][i];
        }
    }
    if (a==b) {
        return a;
    }
    for (int i=k-1; i>=0; i--) {
        if (fel[a][i]!=fel[b][i]) {
            a=fel[a][i], b=fel[b][i];
        }
    }
    return f[a];
}
long long tav(int a, int b) {
    int koz=lca(a, b);
    return dist[a]+dist[b]-2*dist[koz];
}
void keres(int a, int &l, int &r) {
    auto it=s.find(a);
    --it;
    l=*it;
    ++it, ++it;
    if (it!=s.end()) {
        r=*it;
    }
}
void add(int a) {
    s.insert(a);
    int l=0, r=0;
    keres(a, l, r);
    sum+=tav(l, a)+tav(a, r)-tav(l, r);
}
void cl(int a) {
    int l=0, r=0;
    keres(a, l, r);
    sum-=tav(l, a)+tav(a, r)-tav(l, r);
    s.erase(a);
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin >> n >> q;
    p[0]=n+1;
    for (int i=1; i<=n; i++) {
        cin >> p[i];
        f[i]=i-1;
        while (p[i]>p[f[i]]) {
            f[i]=f[f[i]];
        }
        szint[i]=szint[f[i]]+1;
        fel[i][0]=f[i];
        for (int j=1; j<k; j++) {
            fel[i][j]=fel[fel[i][j-1]][j-1];
        }
    }
    for (int i=1; i<=n; i++) {
        cin >> aa[i];
        val[i]+=aa[i];
    }
    for (int i=1; i<=n; i++) {
        cin >> bb[i];
        val[i]-=bb[i];
        val[f[i]]+=bb[i];
    }
    for (int i=n; i>=1; i--) {
        if (val[i]<0) {
            val[f[i]]+=val[i];
            val[i]=0;
        }
    }
    for (int i=1; i<=n; i++) {
        dist[i]=dist[f[i]]+val[i];
    }
    db[0]=1;
    s.insert(0);
    for (int i=1; i<=q; i++) {
        int x; cin >> x;
        if (!v[x]) {
            v[x]=1;
            if (++db[f[x]]==1) {
                add(f[x]);
            }
        } else {
            v[x]=0;
            if (--db[f[x]]==0) {
                cl(f[x]);
            }
        }
        cout << sum/2+val[0] << "\n";
    }
    return 0;
}
