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.
/***************************************************************
	File name: F.cpp
	Author: ljfcnyali
	Create time: 2021年03月18日 星期四 20时57分12秒
***************************************************************/
#include<bits/stdc++.h>

using namespace std;

#define REP(i, a, b) for ( int i = (a), _end_ = (b); i <= _end_; ++ i ) 
#define mem(a) memset ( (a), 0, sizeof ( a ) ) 
#define str(a) strlen ( a ) 
#define pii pair<int, int>
#define int long long
typedef long long LL;

const int maxn = 2e5 + 10;
const int INF = 1e18;

int n, m, q, Q[maxn], p[maxn], a[maxn], b[maxn], dis[maxn], c[maxn], s[maxn], d[maxn];
map<int, int> Map[maxn];

inline int lowbit(int x) { return x & -x; } 
inline void add(int pos, int val) { for ( int i = pos; i <= n + 1; i += lowbit(i) ) c[i] += val; } 
inline int query(int pos) { int ret = 0; for ( int i = pos; i > 0; i -= lowbit(i) ) ret += c[i]; return ret; }
inline void	ADD(int pos, int val) { for ( int i = pos; i <= n; i += lowbit(i) ) d[i] = min(d[i], val); } 
inline int QUERY(int pos) { int ret = n + 1; for ( int i = pos; i > 0; i -= lowbit(i) ) ret = min(ret, d[i]); return ret; } 

signed main()
{
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
	scanf("%lld%lld", &n, &q);
	REP(i, 1, n) scanf("%lld", &p[i]);
	REP(i, 1, n) { scanf("%lld", &a[i]); s[i] = s[i - 1] + a[i]; } 
	REP(i, 1, n) scanf("%lld", &b[i]);
	set<int> Set; Set.insert(1); Set.insert(n + 1); Map[1][n + 1] = true;
	REP(i, 1, q)
	{
		scanf("%lld", &Q[i]); if ( Q[i] == 1 ) continue ; 
		auto it = Set.lower_bound(Q[i]);
		if ( *it != Q[i] ) 
		{
			auto itl = it, itr = it; -- itl; 
			Map[*itl][Q[i]] = Map[Q[i]][*itr] = true;
			Set.insert(Q[i]);
		}
		else
		{
			Set.erase(it); it = Set.lower_bound(Q[i]);
			auto itl = it; -- itl; Map[*itl][*it] = true;
		}
	}

	REP(i, 1, n) d[i] = n + 1;
	for ( int i = n; i >= 1; -- i ) 
	{
		int pos = QUERY(n - p[i] + 1); ADD(n - p[i] + 1, i); add(i + 1, a[i]); 
		if ( query(pos) > b[i] ) add(pos, b[i] - query(pos));
		for ( auto it : Map[i] ) Map[i][it.first] = query(it.first);
	}

	Set.clear(); Set.insert(1); Set.insert(n + 1); int ans = Map[1][n + 1];
	REP(i, 1, q)
	{
		if ( Q[i] != 1 ) 
		{
			auto it = Set.lower_bound(Q[i]);
			if ( *it != Q[i] ) 
			{
				auto itl = it, itr = it; -- itl; 
				ans = ans - Map[*itl][*itr] + Map[*itl][Q[i]] + Map[Q[i]][*itr];
				Set.insert(Q[i]);
			}
			else
			{
				auto itl = it, itr = it; -- itl; ++ itr;
				ans = ans + Map[*itl][*itr] - Map[*itl][Q[i]] - Map[Q[i]][*itr];
				Set.erase(it); 
			}
		}
		printf("%lld\n", ans);
	}
    return 0;
}
