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>
using namespace std;
const int N = 200005;
using ll = long long;
int p[N],st[N],tp,f[N][22],dep[N],n,q,cnt[N];
ll a[N],b[N],val[N],dist[N];bool vis[N];
using ll = long long;
set<int> ss;
int lca(int x,int y)
{
	if (dep[x] < dep[y]) swap(x,y);
    for (int i = 20;i >= 0;--i)
        if (dep[f[x][i]] >= dep[y]) x = f[x][i];
    if (x == y) return x;
    for (int i = 20;i >= 0;--i)
        if (f[x][i] != f[y][i]) x = f[x][i],y = f[y][i];
    return f[x][0];
}
ll dis(int x,int y) {return dist[x] + dist[y] - 2 * dist[lca(x,y)];}
ll sum = 0;
void ins(int x)
{
	auto it = ss.insert(x).first;int l = *(--it);++it;++it;
	int r = (it == ss.end() ? *ss.begin() : *it);
	sum += dis(l,x) + dis(x,r) - dis(l,r);
}
void del(int x)
{
	auto it = ss.find(x);int l = *(--it);++it;++it;
	int r = (it == ss.end() ? *ss.begin() : *it);
	sum -= dis(l,x) + dis(x,r) - dis(l,r);ss.erase(--it);
}
int main ()
{
    ios::sync_with_stdio(false);
    cin >> n >> q;
    for (int i = 1;i <= n;i++)
    {
        cin >> p[i];while (tp && p[st[tp]] < p[i]) --tp;
        dep[i] = dep[f[i][0] = st[tp]] + 1;st[++tp] = i;
        for (int j = 1;(1 << j) <= dep[i];j++) f[i][j] = f[f[i][j - 1]][j - 1]; 
    }
    for (int i = 1;i <= n;i++) cin >> a[i],val[i] += a[i];
    for (int i = 1;i <= n;i++) cin >> b[i],val[i] -= b[i],val[f[i][0]] += b[i];
    for (int i = n;i > 0;--i)
        dist[f[i][0]] += min(0ll,dist[i] += val[i]),dist[i] -= min(dist[i],0ll);
    dist[0] += val[0];
    for (int i = 1;i <= n;i++) dist[i] += dist[f[i][0]];
    cnt[0] = 1;ss.insert(0);
    while (q--)
    {
        int x;cin >> x;
        if (vis[x])
        {
            vis[x] = 0;if (!--cnt[f[x][0]]) del(f[x][0]);
        }
        else
        {
            vis[x] = 1;if (!cnt[f[x][0]]++) ins(f[x][0]);
        }
        cout << sum / 2 + dist[0] << endl;
    }
    return 0;
}