Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std;
using ll = long long;
using pll = pair<ll, ll>;
using plll = tuple<ll, ll, ll>;

struct rmq {
	pll T[550505];
	int sz = 1 << 18;

	void init(ll n, ll *A) {
		int i;
		for (i = 0; i < n; i++) {
			T[sz + i] = pll(A[i], i);
		}
		for (i = sz - 1; i; i--) {
			T[i] = max(T[i << 1], T[i << 1 | 1]);
		}
	}

	pll query(ll l, ll r) {
		pll ret(0, 0);
		for (l += sz, r += sz; l <= r; ){
			if (l & 1) ret = max(ret, T[l]);
			if (~r & 1) ret = max(ret, T[r]);
			l = l + 1 >> 1; r = r - 1 >> 1;
		}
		return ret;
	}
} RMQ;

struct segtree {
	ll T[550505];
	ll sz = 1 << 18;

	void add(ll l, ll r, ll v) {
		for (l += sz, r += sz; l <= r; ) {
			if (l & 1) T[l] += v;
			if (~r & 1) T[r] += v;
			l = l + 1 >> 1; r = r - 1 >> 1;
		}
	}

	ll get(ll i) {
		ll s = 0;
		for (i += sz; i; i >>= 1) {
			s += T[i];
		}
		return s;
	}
} L, R;

set <ll> S;
vector <plll> V[202020];
ll A[202020], B[202020], P[202020], Z[202020];
ll n, q;

void add(ll l, ll r, ll t) {
	V[RMQ.query(l, r).second].emplace_back(l, r, t);
}

void dnc(ll l, ll r) {
	if (l > r) return;
	int m = RMQ.query(l, r).second;
	dnc(l, m - 1); dnc(m + 1, r);

	for (auto &[s, e, t]: V[m]) {
		ll v = R.get(s) + L.get(e);
		if (e > m) v += A[m];
		//cout << s << " " << e << " " << t << ": " << v << endl;
		if (t < 0) Z[-t] -= v;
		else Z[t] += v;
	}

	ll ml = R.get(l);
	ll mr = min(A[m] + R.get(m + 1), B[m]);

	ll lv = ml + A[m];
	ll rv = mr;

	L.add(m, m, ml);
	R.add(m, m, mr);
	L.add(m + 1, r, lv);
	R.add(l, m - 1, rv);

	// cout << "<" << l << " " << r << ">" << ": " << ml << " " << mr << endl;
	// for (int i = l; i <= r; i++) {
	// 	cout << L.get(i) << " / " << R.get(i) << ", ";
	// }
	// cout << endl;
}

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);

	ll i, x;

	cin >> n >> q;

	P[0] = n + 1; P[n + 1] = n + 2;
	for (i = 1; i <= n; i++) {
		cin >> P[i];
	}
	for (i = 1; i <= n; i++) {
		cin >> A[i];	
	}
	for (i = 1; i <= n; i++) {
		cin >> B[i];
	}

	n += 2;

	RMQ.init(n, P);

	add(0, n - 1, 0);

	S.insert(0); S.insert(n - 1);
	for (i = 1; i <= q; i++) {
		cin >> x;
		auto it = S.lower_bound(x);
		if (it == S.end() || *it != x) {
			add(*prev(it), *it, -i);
			add(*prev(it), x, i);
			add(x, *it, i);
			S.insert(x);
		}
		else {
			add(*prev(it), x, -i);
			add(x, *next(it), -i);
			add(*prev(it), *next(it), i);
			S.erase(it);
		}
	}

	dnc(0, n - 1);

	for (i = 1; i <= q; i++) {
		Z[i] += Z[i - 1];
		cout << Z[i] << "\n";
	}

	return 0;
}