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 cs const
#define pb push_back
using namespace std;
typedef long long ll;

cs int N = 2e5 + 50; 
int n, m, p[N]; 
int fa[N][20], dep[N];
int a[N], b[N];
ll dp[N], dlt[N], w[N];
set<int> S; 
ll ans;
int LCA(int x, int y){
	if(dep[x] < dep[y]) swap(x, y);
	for(int i = 18; ~i; i--)
	if(~fa[x][i] && dep[fa[x][i]] >= dep[y]) x = fa[x][i];
	if(x == y) return x; 
	for(int i = 18; ~i; i--)
	if(fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
	return fa[x][0];	
}
ll D(int x, int y){
	return dlt[x] + dlt[y] - 2 * dlt[LCA(x, y)];
}
void ins(int x){
	auto z = S.insert(x).first;
	auto l = z, r = z; --l, ++r; 
	if(r == S.end()) r = S.begin();
	ans += D(*l, x) + D(x, *r) - D(*l, *r);
}
void del(int x){
	auto z = S.find(x);
	auto l = z, r = z; --l, ++r; 
	if(r == S.end()) r = S.begin();
	ans -= D(*l, x) + D(x, *r) - D(*l, *r);
	S.erase(z);
}
int main(){
	#ifdef FSYo
	freopen("1.in", "r", stdin);
	#endif
	cin >> n >> m;
	static int stk[N]; int top = 0; 
	memset(fa, -1, sizeof fa);
	for(int i = 1; i <= n; i++){
		scanf("%d", &p[i]);
		while(top && p[i] > p[stk[top]]) -- top; 
		fa[i][0] = stk[top];
		stk[++top] = i; 
		dep[i] = dep[fa[i][0]] + 1; 
		for(int j = 1; j <= 18; j++)
		if(~fa[i][j - 1]) fa[i][j] = fa[fa[i][j - 1]][j - 1];
	}
	for(int i = 1; i <= n; i++)
		scanf("%d", &a[i]);
	for(int i = 1; i <= n; i++)
		scanf("%d", &b[i]);
	for(int i = 1; i <= n; i++){
		w[i] += a[i] - b[i];
		w[fa[i][0]] += b[i];
	}
	for(int i = n; i; i--){
		dp[i] += w[i];
		dp[fa[i][0]] += min(0ll, dp[i]);
	}
	ll cur = dp[0] + w[0];
	static int ct[N]; ct[0] = 1, S.insert(0);
	for(int i = 1; i <= n; i++)
		dlt[i] = max(0ll, dp[i]) + dlt[fa[i][0]];
	static bool o[N];
	while(m--){
		int x; scanf("%d", &x);
		int f = fa[x][0];
		if(!o[x]) {
			if(!ct[f]) ins(f);
			++ ct[f];
		}
		else {
			-- ct[f];
			if(!ct[f]) del(f);
		} o[x] ^= 1;
		cout << ans / 2 + cur << '\n';
	} return 0; 
}