Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

typedef long long ll;

const int N = 200000;
const int S = 80*N;

int ch[2][S + 5], ncnt; ll tg[S + 5];

ll query(int x, int l, int r, int p) {
	if( !x ) return 0;
	
	int m = (l + r) >> 1;
	if( p <= m ) return query(ch[0][x], l, m, p) + tg[x];
	else return query(ch[1][x], m + 1, r, p) + tg[x];
}
int add(int pre, int l, int r, int p, ll k) {
	if( p < l ) return pre;
	
	int x = (++ncnt);
	ch[0][x] = ch[0][pre], ch[1][x] = ch[1][pre], tg[x] = tg[pre];
	if( r <= p ) {tg[x] += k; return x;}
	else {
		int m = (l + r) >> 1;
		ch[0][x] = add(ch[0][x], l, m, p, k);
		ch[1][x] = add(ch[1][x], m + 1, r, p, k);
		return x;
	}
}

int a[N + 5], b[N + 5], p[N + 5], rt[N + 5], n, q;

int stk[N + 5], tp;
int main() {
	scanf("%d%d", &n, &q);
	for(int i=1;i<=n;i++) scanf("%d", &p[i]); p[n + 1] = n + 1;
	for(int i=1;i<=n;i++) scanf("%d", &a[i]);
	for(int i=1;i<=n;i++) scanf("%d", &b[i]);
	
	stk[++tp] = 1;
	for(int i=2;i<=n+1;i++) {
		rt[i] = add(rt[i - 1], 1, n, i - 1, a[i - 1]);
		
		while( tp && p[i] > p[stk[tp]] ) {
			int j = stk[tp--]; ll k = query(rt[i], 1, n, j);
			if( k > b[j] ) rt[i] = add(rt[i], 1, n, j, b[j] - k);
		}
		stk[++tp] = i;
	}
	
	ll ans = query(rt[n + 1], 1, n, 1);
	std::set<int>st; st.insert(1), st.insert(n + 1);
	for(int i=1,x;i<=q;i++) {
		scanf("%d", &x);
		if( x != 1 ) {
			if( st.count(x) ) {
				auto it = st.find(x);
				
				int l = -1, r = -1;
				it++; if( it != st.end() ) r = *it; it--;
				if( it != st.begin() ) it--, l = *it, it++;
				assert( l != -1 && r != -1 );
				
				ans -= query(rt[x], 1, n, l) + query(rt[r], 1, n, x) - query(rt[r], 1, n, l);
				
				st.erase(x);
			} else {
				auto it = st.insert(x).first;
				
				int l = -1, r = -1;
				it++; if( it != st.end() ) r = *it; it--;
				if( it != st.begin() ) it--, l = *it, it++;
				assert( l != -1 && r != -1 );
				
				ans += query(rt[x], 1, n, l) + query(rt[r], 1, n, x) - query(rt[r], 1, n, l);
			}
		}
		printf("%lld\n", ans);
	}
}