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 <cmath>
#include <vector>
#include <cstdio>
#include <cstddef>
#include <cstring>
#include <iostream>
#include <algorithm>
#define pii pair<int, int>
#define mp make_pair
#define int long long
using namespace std;
 
inline int read() {
	int x = 0, w = 1;char ch = getchar();
	while (ch > '9' || ch < '0') { if (ch == '-')w = -1;ch = getchar(); }
	while (ch >= '0' && ch <= '9')x = x * 10 + ch - '0', ch = getchar();
	return x * w;
}
inline void write(int x) {
    if (x < 0) putchar('-'), x = -x;
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
}
 
const int maxn = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f3f3f3f;
 
inline int min(int x, int y) { return x < y ? x : y; }
inline int max(int x, int y) { return x > y ? x : y; }

struct node {
	int nxt, to;
} edge[maxn << 1];
int n, q, top, tot, head[maxn], st[maxn], dep[maxn], fa[maxn][25];
int a[maxn], b[maxn], val[maxn], dis[maxn], p[maxn];

inline void addedge(int u, int v) {
	edge[++tot] = (node){head[u], v}, head[u] = tot;
}
inline int getlca(int u, int v) {
	if (dep[u] > dep[v]) swap(u, v);
	for (int i = 20;i >= 0;i--)
		if (dep[u] <= dep[v] - (1 << i)) v = fa[v][i];
	if (u == v) return u;
	for (int i = 20;i >= 0;i--)
		if (fa[u][i] != fa[v][i])
			u = fa[u][i], v = fa[v][i];
	return fa[u][0];
}

int sum, cnt[maxn], vis[maxn]; set<int> s;
inline int getdis(int x, int y) {
	return dis[x] + dis[y] - (dis[getlca(x, y)] << 1);
}
set<int> :: iterator it;
inline void insert(int x) {
	it = s.insert(x).first;
	int l = *(--it); ++it;++it;
	int r = ((it == s.end()) ? *s.begin() : *it);
	sum += getdis(l, x) + getdis(x, r) - getdis(l, r);
}
inline void remove(int x) {
	it = s.find(x);
	int l = *(--it); ++it;++it;
	int r = ((it == s.end()) ? *s.begin() : *it);
	sum -= getdis(l, x) + getdis(x, r) - getdis(l, r);
	--it; s.erase(it);
}

signed main(void) {
	n = read(), q = read();
	for (int i = 1;i <= n;i++) {
		p[i] = read();
		while (top && p[st[top]] < p[i]) top--;
		fa[i][0] = st[top], dep[i] = dep[st[top]] + 1, st[++top] = i;
		for (int j = 1;(1 << j) <= dep[i];j++)
			fa[i][j] = fa[fa[i][j - 1]][j - 1];
	}
	for (int i = 1;i <= n;i++) a[i] = read(), val[i] += a[i];
	for (int i = 1;i <= n;i++) b[i] = read(), val[i] -= b[i], val[fa[i][0]] += b[i];
	for (int i = n;i >= 1;i--) {
		dis[fa[i][0]] += min(0, dis[i] + val[i]);
		dis[i] += val[i], dis[i] -= min(0, dis[i]);
	}
	dis[0] += val[0];
	for (int i = 1;i <= n;i++) dis[i] += dis[fa[i][0]];
	cnt[0] = 1, s.insert(0);
	while (q--) {
		int x = read();
		if (vis[x]) { 
			vis[x] = 0;
			if (!(--cnt[fa[x][0]])) remove(fa[x][0]);
		} else {
			vis[x] = 1;
			if (!(cnt[fa[x][0]]++)) insert(fa[x][0]);
 		}
		printf("%lld\n", (sum >> 1) + dis[0]);
	}
    return 0; 
}