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.
/**
 *  created: 31/03/2021, 23:09:03
**/

#include <bits/stdc++.h>

using namespace std;

const int max_n = 200222, inf = 1000111222;
const int max_lev = 18;

int n, q, p[max_n], a[max_n], nxt[max_n];
int fin[max_lev][max_n];
long long cost[max_lev][max_n], b[max_n];

void calc_nxt() {
    p[n] = inf;
    nxt[n] = n;
    for (int i = n - 1; i >= 0; --i) {
        nxt[i] = i + 1;
        while (p[i] > p[nxt[i]]) {
            nxt[i] = nxt[nxt[i]];
        }
    }
}

void calc_fin() {
    for (int i = 0; i < max_lev; ++i) {
        fin[i][n] = n;
    }
    for (int i = n - 1; i >= 0; --i) {
        nxt[i] = i + 1;
        while (p[i] > p[nxt[i]]) {
            nxt[i] = nxt[nxt[i]];
        }
        long long tot = a[i];
        int pos = i + 1;
        for (int lev = max_lev - 1; lev >= 0; --lev) {
            if (fin[lev][pos] <= nxt[i]) {
                tot += cost[lev][pos];
                pos = fin[lev][pos];
            }
        }
        b[i] = min(b[i], tot);
        fin[0][i] = nxt[i];
        cost[0][i] = b[i];
        for (int lev = 1; lev < max_lev; ++lev) {
            fin[lev][i] = fin[lev - 1][fin[lev - 1][i]];
            cost[lev][i] = cost[lev - 1][i] + cost[lev - 1][fin[lev - 1][i]];
        }
    }
}

const int magic = 4444;

int light_fin[max_lev][max_n], mn[max_lev][max_n];
long long light_cost[max_lev][max_n];

void calc_light_nxt() {
    for (int lev = 0; lev < max_lev; ++lev) {
        light_fin[lev][n] = n;
        mn[lev][n] = inf;
    }
    for (int i = n - 1; i >= 0; --i) {
        mn[0][i] = inf;
        int light_nxt = nxt[i];
        long long light_a = b[i];
        if (nxt[i] - i > magic) {
            light_nxt = i + 1;
            light_a = a[i];
            mn[0][i] = nxt[i];
        }

        light_fin[0][i] = light_nxt;
        light_cost[0][i] = light_a;
        for (int lev = 1; lev < max_lev; ++lev) {
            light_fin[lev][i] = light_fin[lev - 1][light_fin[lev - 1][i]];
            light_cost[lev][i] = light_cost[lev - 1][i] + light_cost[lev - 1][light_fin[lev - 1][i]];
            mn[lev][i] = min(mn[lev - 1][i], mn[lev - 1][light_fin[lev - 1][i]]);
        }
    }
}

map<pair<int, int>, long long> rem;

long long get_cost(int from, int to) {
    if (rem.count({from, to})) {
        return rem[{from, to}];
    }
    long long res = 0;
    //cout << from << " -> " << to << endl;
    while (to - from > magic) {
        //cout << from << ": " << light_fin[0][from] << " " << light_cost[0][from] << " " << mn[0][from] << endl;
        for (int lev = max_lev - 1; lev >= 0; --lev) {
            if (light_fin[lev][from] > to) {
                continue;
            }
            if (mn[lev][from] > to) {
                //cout << "#" << lev << " " << light_fin[lev][from] << endl;
                res += light_cost[lev][from];
                from = light_fin[lev][from];
            }
        }
        if (nxt[from] <= to) {
            res += b[from];
            from = nxt[from];
        }
    }
    while (from < to) {
        if (nxt[from] <= to) {
            res += b[from];
            from = nxt[from];
        } else {
            res += a[from];
            ++from;
        }
    }
    return rem[{from, to}] = res;
}

int main() {
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    scanf("%d%d", &n, &q);
    for (int i = 0; i < n; ++i) {
        scanf("%d", &p[i]);
    }
    for (int i = 0; i < n; ++i) {
        scanf("%d", &a[i]);
    }
    for (int i = 0; i < n; ++i) {
        scanf("%lld", &b[i]);
    }
    calc_nxt();
    calc_fin();
    calc_light_nxt();
    set<int> all{0, n};
    long long ans = get_cost(0, n);
    while (q--) {
        int x;
        scanf("%d", &x);
        --x;
        if (x) {
            auto p = all.insert(x);
            auto it = p.first;
            if (!p.second) {
                int pr = *prev(it);
                int nxt = *next(it);
                ans -= get_cost(pr, x);
                ans -= get_cost(x, nxt);
                ans += get_cost(pr, nxt);
                all.erase(it);
            } else {
                int pr = *prev(it);
                int nxt = *next(it);
                ans += get_cost(pr, x);
                ans += get_cost(x, nxt);
                ans -= get_cost(pr, nxt);
            }
        }
        printf("%lld\n", ans);
    }
    return 0;
}
