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;

#define sz(x) ((int) ((x).size()))
typedef long long ll;
typedef long double ld;

int n, q, p[200100];
ll a[200100], b[200100];

int m[2][200100][18];
ll c[2][200100][18];

void computeTable(int k, int i) {
  for (int b = 1; b < 18; b++) {
    m[k][i][b] = m[k][m[k][i][b - 1]][b - 1];
    c[k][i][b] = c[k][i][b - 1] + c[k][m[k][i][b - 1]][b - 1];
  }
}

ll cost(int k, int i, int j) {
  ll r = 0;
  for (int b = 17; b >= 0; b--)
    if (k == 0 && m[k][i][b] <= j || k == 1 && m[k][i][b] >= j)
      r += c[k][i][b], i = m[k][i][b];
  return r;
}

ll cost(int i, int j) {
  return cost(0, i, j) + cost(1, j, i);
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(10);
  cin >> n >> q;
  for (int i = 0; i < n; i++)
    cin >> p[i], p[i]--;
  p[n] = n;
  for (int i = 0; i < n; i++)
    cin >> a[i];
  for (int i = 0; i < n; i++)
    cin >> b[i];
  stack<int> s;
  s.push(n);
  m[0][n][0] = n;
  c[0][n][0] = 0;
  computeTable(0, n);
  for (int i = n - 1; i >= 0; i--) {
    while (!s.empty() && p[s.top()] < p[i])
      s.pop();
    if (s.empty()) {
      m[0][i][0] = i;
      c[0][i][0] = 0;
    } else {
      int j = s.top();
      m[0][i][0] = j;
      c[0][i][0] = min(b[i], a[i] + cost(0, i + 1, j));
    }
    computeTable(0, i);
    s.push(i);
  }
  while (!s.empty())
    s.pop();
  for (int i = 0; i <= n; i++) {
    while (!s.empty() && p[s.top()] < p[i])
      s.pop();
    if (s.empty()) {
      m[1][i][0] = i;
      c[1][i][0] = 0;
    } else {
      int j = s.top();
      m[1][i][0] = j;
      c[1][i][0] = a[j] + cost(0, j + 1, i);
    }
    computeTable(1, i);
    s.push(i);
  }
  set<int> cs;
  cs.insert(0);
  cs.insert(n);
  ll res = cost(0, n);
  int x;
  for (int i = 0; i < q; i++) {
    cin >> x, x--;
    if (x > 0) {
      int f = 1;
      if (cs.find(x) != cs.end())
        cs.erase(cs.find(x)), f = -1;
      auto it = cs.lower_bound(x);
      int k = *it;
      it--;
      int j = *it;
      res += f * (cost(j, x) + cost(x, k) - cost(j, k));
      if (f == 1)
        cs.insert(x);
    }
    cout << res << "\n";
  }
}
