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 <set>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define N 200005
#define int long long
const int inf = 1e15;
int n, q, P[N], A[N], B[N], fat[N], stk[N], top, lef[N];
int d[N], dep[N], f[N][20], ans[N], col[N], sum[N], rev[N];
int ans2[N], beg[N], mnn[N];
int Get(int x, int y) { return ans[x] - ans[y] ; }
int Get2(int x, int y) { return ans2[x] - ans2[y] ; }
int Getup(int x, int k)
{
	for (int i = 18; i >= 0; i--)
		if (dep[f[x][i]] >= k)
			x = f[x][i];
	return x;
}
int Lca(int x, int y)
{
	if (dep[x] < dep[y])
		swap(x, y);
	for (int i = 18; i >= 0; i--)
		if (dep[f[x][i]] >= dep[y])
			x = f[x][i];
	for (int i = 18; i >= 0; i--)
		if (f[x][i] != f[y][i])
			x = f[x][i], y = f[y][i];
	return (x == y) ? x : f[x][0];
}
set<int>S;
set<int>::iterator it ;
vector<int> G[N];
int now;
int Getdis(int x, int y)
{
	if (x > y)
		swap(x, y);
	int c = Lca(x, y);
	if (c == y)
		return Get(x, y) ;
	int u = Getup(x, dep[c] + 1), v = Getup(y, dep[c] + 1) ;
	if (u > v)
		swap(u, v), swap(x, y);
	int ret = Get(x, u) + rev[v] - sum[u] + Get2(y, v) + A[lef[v]] + Get(mnn[y], y);
	return ret;
}
void Dfs(int x)
{
	mnn[x] = x;
	for (int i = 0; i < (int)G[x].size(); i++)
	{
		int v = G[x][i];
		Dfs(v);
		mnn[x] = min(mnn[x], mnn[v]);
	}
}
signed main()
{
	scanf("%lld%lld", &n, &q);
	for (int i = 1; i <= n; i++)
		scanf("%lld", &P[i]);
	for (int i = 1; i <= n; i++)
		scanf("%lld", &A[i]);
	for (int i = 1; i <= n; i++)
		scanf("%lld", &B[i]);
	P[n + 1] = n + 1;
	stk[++top] = n + 1;
	dep[n + 1] = 1;
	for (int i = n; i >= 1; i--)
	{
		while (P[i] > P[stk[top]])
			top--;
		fat[i] = stk[top];
		G[stk[top]].push_back(i);
		dep[i] = dep[fat[i]] + 1 ;
		d[i] = min(B[i], A[i] + Get(i + 1, fat[i]));
		ans[i] = d[i], f[i][0] = fat[i];
		for (int j = 1; j <= 18; j++)
			f[i][j] = f[f[i][j - 1]][j - 1];
		ans[i] += ans[fat[i]], stk[++ top] = i;
	}
	for (int i = n + 1; i >= 1; i--)
	{
		sort(G[i].begin(), G[i].end()) ; int bef = 0 ;
		for (int j = 0; j < (int)G[i].size(); j++)
		{
			int v = G[i][j];
			if (!bef)
				sum[v] = 0, beg[i] = v ;
			else
			{
				sum[v] = sum[bef] + A[bef] + Get(bef + 1, v);
				rev[v] = sum[bef], lef[v] = bef;
			}
			bef = v;
		}
	}
	Dfs(n + 1);
	for (int i = n; i >= 1; i--)
	{
		int ll = mnn[fat[i]];
		int u = Getup(ll, dep[i]);
		if (i == beg[fat[i]])
			ans2[i] = 0 ;
		else
			ans2[i] = Get(ll, u) + rev[i] - sum[u] + A[lef[i]];
		ans2[i] += ans2[fat[i]];
	}
	S.insert(1);
	S.insert(n + 1);
	now = Get(1, n + 1);
	while (q--)
	{
		int x;
		scanf("%lld", &x);
		col[x] ^= 1 ;
		if (x == 1)
			printf("%lld\n", now);
		else
		{
			if (!col[x])
				S.erase(x);
			int las = 0, nxt = 0;
			it = S.upper_bound(x), nxt = *it;
			--it, las = *it;
			if (col[x])
			{
				now -= Getdis(las, nxt);
				now += Getdis(las, x);
				now += Getdis(x, nxt);
				S.insert(x);
			}
			else
			{
				now += Getdis(las, nxt);
				now -= Getdis(las, x);
				now -= Getdis(x, nxt);
			}
			printf("%lld\n", now);
		}
	}
}
