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;
const int LOGN = 18;
int n, q, p[N], stk[N], L[LOGN][N], R[LOGN][N];
long long a[N], b[N], LC[LOGN][N], RC[LOGN][N];
inline long long getdist(int u, int v) {
	long long ret = 0;
	for (int i = LOGN - 1; i >= 0; i--) if (R[i][u] <= v) {
		ret += RC[i][u];
		u = R[i][u];
	}
	if (u == v) return ret;
	for (int i = LOGN - 1; i >= 0; i--) if (L[i][v] >= u) {
		ret += LC[i][v];
		v = L[i][v];
	}
	return ret;
}
int main() {
	scanf("%d%d", &n, &q);
	for (int i = 1; i <= n; i++) scanf("%d", &p[i]);
	p[n + 1] = n + 1;
	for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
	for (int i = 1; i <= n; i++) scanf("%lld", &b[i]);
	int tp = 0;
	stk[++tp] = n + 1;
	for (int j = 0; j < LOGN; j++) {
		R[j][n + 1] = n + 1;
		RC[j][n + 1] = 0;
	}
	for (int i = n; i >= 1; i--) {
		while (tp && p[stk[tp]] < p[i]) tp--;
		R[0][i] = stk[tp];
		RC[0][i] = b[i] = min(b[i], a[i] + getdist(i + 1, stk[tp]));
		for (int j = 1; j < LOGN; j++) {
			R[j][i] = R[j - 1][R[j - 1][i]];
			RC[j][i] = RC[j - 1][i] + RC[j - 1][R[j - 1][i]];
		}
		stk[++tp] = i;
	}
	tp = 0;
	stk[++tp] = 1;
	for (int j = 0; j < LOGN; j++) {
		L[j][1] = L[j][0] = 0;
		LC[j][1] = 0;
	}
	for (int i = 2; i <= n + 1; i++) {
		while (tp && p[stk[tp]] < p[i]) tp--;
		L[0][i] = tp ? stk[tp] : 0;
		LC[0][i] = tp ? a[stk[tp]] + getdist(stk[tp] + 1, i) : 0;
		for (int j = 1; j < LOGN; j++) {
			L[j][i] = L[j - 1][L[j - 1][i]];
			LC[j][i] = LC[j - 1][i] + LC[j - 1][L[j - 1][i]];
		}
		stk[++tp] = i;
	}
	set<int> st;
	st.insert(1);
	st.insert(n + 1);
	long long sum = getdist(1, n + 1);
	while (q--) {
		int x;
		scanf("%d", &x);
		if (x != 1) {
			if (st.count(x)) {
				auto it = st.find(x);
				int prv = *prev(it), nxt = *next(it);
				sum += getdist(prv, nxt);
				sum -= getdist(prv, x);
				sum -= getdist(x, nxt);
				st.erase(it);
			} else {
				auto it = st.insert(x).first;
				int prv = *prev(it), nxt = *next(it);
				sum -= getdist(prv, nxt);
				sum += getdist(prv, x);
				sum += getdist(x, nxt);
			}
		}
		printf("%lld\n", sum);
	}
	return 0;
}