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;

typedef long long ll;
const int maxn = 200010;
int n, q, p[maxn], dep[maxn], fa[18][maxn], in[maxn], cnt[maxn];
ll a[maxn], b[maxn], c[maxn], d[maxn];

int main() {
    scanf("%d %d", &n, &q);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &p[i]);
    }
    for (int i = 1; i <= n; i++) {
        scanf("%lld", &a[i]);
    }
    for (int i = 1; i <= n; i++) {
        scanf("%lld", &b[i]), c[i] = a[i] - b[i];
    }
    stack<int> st; st.push(0);
    for (int i = 1; i <= n; i++) {
        while (st.size() > 1 && p[st.top()] < p[i]) st.pop();
        dep[i] = dep[fa[0][i] = st.top()] + 1;
        c[st.top()] += b[i], st.push(i);
    }
    for (int i = n; i; i--) {
        d[fa[0][i]] += min(0LL, d[i] += c[i]);
    }
    d[0] += c[0];
    for (int i = 1; i <= n; i++) {
        d[i] += d[fa[0][i]] - min(0LL, d[i]);
    }
    for (int i = 1; i <= 17; i++) {
        for (int j = 0; j <= n; j++) {
            fa[i][j] = fa[i - 1][fa[i - 1][j]];
        }
    }
    auto lca = [&](int u, int v) {
        if (dep[u] < dep[v]) swap(u, v);
        for (int i = 17; ~i; i--) {
            if (dep[fa[i][u]] >= dep[v]) u = fa[i][u];
        }
        if (u == v) return u;
        for (int i = 17; ~i; i--) {
            if (fa[i][u] ^ fa[i][v]) u = fa[i][u], v = fa[i][v];
        }
        return fa[0][u];
    };
    set<int> S = {0};
    ll ans = d[0];
    auto solve = [&](int x, int coef) {
        if (!x) return;
        auto pre = S.lower_bound(x), nxt = S.upper_bound(x);
        int u = pre == S.begin() ? -1 : *--pre;
        int v = nxt == S.end() ? -1 : *nxt;
        if (~u && ~v) ans += coef * d[lca(u, v)];
        if (~u) ans -= coef * d[lca(u, x)];
        if (~v) ans -= coef * d[lca(v, x)];
        ans += coef * d[x];
        coef == 1 ? (void)S.insert(x) : (void)S.erase(x);
    };
    for (int _ = 1, x; _ <= q; _++) {
        scanf("%d", &x);
        if (in[x]) {
            if (!--cnt[fa[0][x]]) solve(fa[0][x], -1);
        } else {
            if (!cnt[fa[0][x]]++) solve(fa[0][x], 1);
        }
        in[x] ^= 1, printf("%lld\n", ans);
    }
    return 0;
}