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>
const int N = 2e5 + 10, P = 1e9 + 7;
typedef long long LL;
int ri() {
    int x = 0, f = 1;
    char ch = getchar();
    for(;ch < '0' || ch > '9'; ch = getchar())
        if(ch == '-')
            f = -1;
    for(;ch >= '0' && ch <= '9'; ch = getchar())
        x = (x << 1) + (x << 3) + ch - '0';
    return x * f;
}
int n, q, p[N], fa[22][N], de[N], cnt[N];
std::set<int>S;
std::set<int>::iterator it;
LL val[N], f[N], Sum;
int lca(int u, int v) {
    if(de[u] > de[v]) std::swap(u, v);
    for(int i = 20; ~i; --i)
        if(de[fa[i][v]] >= de[u])
            v = fa[i][v];
    if(u == v) return u;
    for(int i = 20; ~i; --i)
        if(fa[i][u] != fa[i][v])
            u = fa[i][u], v = fa[i][v];
    return fa[0][u];
}
LL Dis(int u, int v) {return f[u] + f[v] - 2 * f[lca(u, v)];}
void del(int x) {
    it = S.erase(S.find(x));
    int r, l;
    if(it == S.end()) r = *S.begin();
    else r = *it;
    l = *--it;
    Sum -= Dis(l, x) + Dis(x, r) - Dis(l, r);
}
void ins(int x) {
    it = S.insert(x).first;
    int l = *--it, r;
    ++it; ++it;
    if(it == S.end()) r = *S.begin();
    else r = *it;
    Sum += Dis(l, x) + Dis(x, r) - Dis(l, r);
}
int main() {
    n = ri(); q = ri();
    static int st[N], tp = 0;
    st[0] = 0;
    for(int i = 1;i <= n; ++i) {
        p[i] = ri();
        for(;tp && p[st[tp]] < p[i]; --tp) ;
        fa[0][i] = st[tp];
        de[i] = de[st[tp]] + 1;
        st[++tp] = i;
        for(int j = 1;j <= 20; ++j)
            fa[j][i] = fa[j - 1][fa[j - 1][i]];
    }
    for(int i = 1;i <= n; ++i)
        val[i] += ri();
    for(int i = 1;i <= n; ++i) {
        int b = ri();
        val[i] -= b; val[fa[0][i]] += b;
    }

    for(int i = n;i; --i) {
        f[i] += val[i];
        if(f[i] < 0)
            f[fa[0][i]] += f[i], f[i] = 0;
    }

    f[0] += val[0];
    for(int i = 1;i <= n; ++i)
        f[i] += f[fa[0][i]];
    cnt[0] = 1; S.insert(0);
    static bool vis[N] = {0};
    for(;q--;) {
        int x = ri();
        if(vis[x] == true) {
            vis[x] = false;
            if(!--cnt[fa[0][x]])
                del(fa[0][x]);
        }
        else {
            vis[x] = true;
            if(!cnt[fa[0][x]]++)
                ins(fa[0][x]);
        }
        printf("%lld\n", Sum / 2 + f[0]);
    }
    return 0;
}
/*
 100
1
100 2 0 1
 */