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>

#define LL long long
#define ull unsigned long long
#define F(i, j, k) for (int i = j; i <= k; ++i)
#define DF(i, j, k) for (int i = j; i >= k; --i)

using namespace std;

template <typename T> inline void read(T &n) {
    T w = 1; n = 0;
    char ch = getchar();
    while (!isdigit(ch) && ch != EOF) { if (ch == '-') w = -1; ch = getchar(); }
    while (isdigit(ch) && ch != EOF) { n = (n << 3) + (n << 1) + (ch & 15); ch = getchar(); }
    n *= w;
}

template <typename T> inline void write(T x) {
    T l = 0; ull y = 0;
    if (!x) { putchar(48); return; }
    if (x < 0) { x = -x; putchar('-'); }
    while (x) { y = y * 10 + x % 10; x /= 10; ++l; }
    while (l) { putchar(y % 10 + 48); y /= 10; --l; }
}

template <typename T> inline void writes(T x) {
    write(x);
    putchar(' ');
}

template <typename T> inline void writeln(T x) {
    write(x);
    puts("");
}

template <typename T> inline void checkmax(T &a, T b) { a = a > b ? a : b; }

template <typename T> inline void checkmin(T &a, T b) { a = a < b ? a : b; }

const int N = 2e5 + 100, K = 20;

int n, q, p[N], a[N], b[N], st[N], top, fa[N][K], depth[N], num[N];

bool vis[N];

LL c[N], ans;

set <int> S;

inline int lca(int x, int y) {
    if (depth[x] > depth[y]) swap(x, y);
    DF(i, 18, 0) if (depth[fa[y][i]] >= depth[x]) y = fa[y][i];
    if (x == y) return x;
    DF(i, 18, 0) if (fa[x][i] != fa[y][i]) { x = fa[x][i]; y = fa[y][i]; }
    return fa[x][0];
}

inline LL dist(int x, int y) { return c[x] + c[y] - (c[lca(x, y)] << 1); }

inline void del(int x) {
    S.erase(x);
    set <int> :: iterator it = S.lower_bound(x);
    int nxt, fro;
    if (it == S.end()) {
        nxt = *S.begin();
        fro = *(--it);
    }
    else  {
        nxt = *it;
        fro = *(--it);
    }
    ans -= dist(x, fro) + dist(x, nxt) - dist(fro, nxt);
}

inline void ins(int x) {
    set <int> :: iterator it = S.lower_bound(x);
    int nxt, fro;
    if (it == S.end()) {
        nxt = *S.begin();
        fro = *(--it);
    }
    else  {
        nxt = *it;
        fro = *(--it);
    }
    ans += dist(x, fro) + dist(x, nxt) - dist(fro, nxt);
    S.insert(x);
}

int main () {
    //freopen(".in", "r", stdin);
    //freopen(".out", "w", stdout);
    read(n);
    read(q);
    F(i, 1, n) read(p[i]);
    F(i, 1, n) read(a[i]);
    F(i, 1, n) read(b[i]);
    depth[0] = 1;
    F(i, 1, n) {
        while (top && p[st[top]] < p[i]) --top;
        fa[i][0] = st[top];
        depth[i] = depth[fa[i][0]] + 1;
        st[++top] = i;
    }
    F(i, 1, 18)
        F(j, 1, n) fa[j][i] = fa[fa[j][i - 1]][i - 1];
    F(i, 1, n) { c[i] = a[i] - b[i]; c[fa[i][0]] += b[i]; }
    DF(i, n, 1)
        if (c[i] < 0) {
            c[fa[i][0]] += c[i];
            c[i] = 0;
        }
    F(i, 1, n) c[i] += c[fa[i][0]];
    num[0] = 1;
    S.insert(0);
    F(i, 1, q) {
        int x;
        read(x);
        if (vis[x]) {
            --num[fa[x][0]];
            if (!num[fa[x][0]]) del(fa[x][0]);
            vis[x] = 0;
        }
        else {
            if (!num[fa[x][0]]) ins(fa[x][0]);
            ++num[fa[x][0]];
            vis[x] = 1;
        }
        writeln((ans >> 1) + c[0]);
    }
    return 0;
}