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:
/*Lucky_Glass*/
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

inline int rin(int &r) {
	int b = 1, c = getchar(); r = 0;
	while( c < '0' || '9' < c ) b = c == '-' ? -1 : b, c = getchar();
	while( '0' <= c && c <= '9' ) r = (r * 10) + (c ^ '0'), c = getchar();
	return r *= b;
}

const int N = 2e5 + 10;
typedef long long llong;

int n, m, nstk;
llong ans;
int arp[N], ara[N], arb[N], stk[N], fa[N][20], dep[N];
llong arc[N], dp[N], sum[N], cnt[N];
set<int> fix;
bool vis[N];

void build() {
	nstk = 1; arp[0] = n + 1;
	for(int i = 1; i <= n; i++) {
		while( nstk && arp[stk[nstk]] < arp[i] ) nstk--;
		fa[i][0] = stk[nstk];
		arc[fa[i][0]] += arb[i];
		if( !fa[i][0] ) ans += arb[i];
		dep[i] = dep[fa[i][0]] + 1;
		for(int j = 1; j < 20; j++)
			fa[i][j] = fa[fa[i][j - 1]][j - 1];
		stk[++nstk] = i;
	}
}
int lca(int u, int v) {
	if( dep[u] < dep[v] ) swap(u, v);
	for(int i = 19; ~i; i--)
		if( dep[fa[u][i]] >= dep[v] )
			u = fa[u][i];
	if( u == v ) return u;
	for(int i = 19; ~i; i--)
		if( fa[u][i] != fa[v][i] )
			u = fa[u][i], v = fa[v][i];
	return fa[u][0];
}
void ina_remove(int x) {
	set<int>::iterator it = fix.find(x);
	int llca = 0, rlca = 0;
	if( it != fix.begin() ) llca = lca(*(--it), x), it++;
	if( (++it) != fix.end() ) rlca = lca(*it, x);
	it--;
	fix.erase(it);
	if( dep[llca] < dep[rlca] ) llca = rlca;
	ans -= sum[x] - sum[llca];
}
void ina_resume(int x) {
	fix.insert(x);
	set<int>::iterator it = fix.find(x);
	int llca = 0, rlca = 0;
	if( it != fix.begin() ) llca = lca(*(--it), x), it++;
	if( (++it) != fix.end() ) rlca = lca(*it, x);
	if( dep[llca] < dep[rlca] ) llca = rlca;
	ans += sum[x] - sum[llca];
}
int main() {
	rin(n), rin(m);
	for(int i = 1; i <= n; i++) rin(arp[i]);
	for(int i = 1; i <= n; i++) rin(ara[i]), arc[i] += ara[i];
	for(int i = 1; i <= n; i++) rin(arb[i]), arc[i] -= arb[i];
	build();
	// dp[i] : the min extra-cost to choose i when already choose fa[i][0]
	for(int i = n; i; i--) {
		dp[i] += arc[i];
		if( dp[i] < 0 ) dp[fa[i][0]] += dp[i];
	}
	dp[0] += arc[0];
	// sum[i] : the min cost to choose points from root to i
	sum[0] = dp[0];
	for(int i = 1; i <= n; i++)
		if( dp[i] < 0 ) // i is already chosen when choose fa[i][0]
			sum[i] = sum[fa[i][0]];
		else
			sum[i] = sum[fa[i][0]] + dp[i];
	fix.insert(0), cnt[0] = 1, ans = sum[0];
	for(int i = 1, s; i <= m; i++) {
		rin(s);
		if( vis[s] ) {
			vis[s] = false;
			if( !(--cnt[fa[s][0]]) ) ina_remove(fa[s][0]);
		}
		else {
			vis[s] = true;
			if( !(cnt[fa[s][0]]++) ) ina_resume(fa[s][0]);
		}
		printf("%lld\n", ans);
	}
	return 0;
}