[数据结构与算法]E - Yet Another Card Deck

?E - Yet Another Card Deck

You have a card deck of?nn?cards, numbered from top to bottom, i.?e. the top card has index?11?and bottom card?— index?nn. Each card has its color: the?ii-th card has color?a_iai?.

You should process?qq?queries. The?jj-th query is described by integer?t_jtj?. For each query you should:

• find the highest card in the deck with color?t_jtj?, i.?e. the card with minimum index;
• print the position of the card you found;
• take the card and place it on top of the deck.

Input

The first line contains two integers?nn?and?qq?(2 \le n \le 3 \cdot 10^52≤n≤3?105;?1 \le q \le 3 \cdot 10^51≤q≤3?105)?— the number of cards in the deck and the number of queries.

The second line contains?nn?integers?a_1, a_2, \dots, a_na1?,a2?,…,an??(1 \le a_i \le 501≤ai?≤50)?— the colors of cards.

The third line contains?qq?integers?t_1, t_2, \dots, t_qt1?,t2?,…,tq??(1 \le t_j \le 501≤tj?≤50)?— the query colors. It's guaranteed that?queries ask only colors that are present in the deck.

Output

Print?qq?integers?— the answers for each query.

Example

Input

7 5
2 1 1 4 3 3 1
3 2 1 1 4

Output

5 2 3 1 5 

Note

Description of the sample:

1. the deck is?[2, 1, 1, 4, \underline{3}, 3, 1][2,1,1,4,3?,3,1]?and the first card with color?t_1 = 3t1?=3?has position?55;
2. the deck is?[3, \underline{2}, 1, 1, 4, 3, 1][3,2?,1,1,4,3,1]?and the first card with color?t_2 = 2t2?=2?has position?22;
3. the deck is?[2, 3, \underline{1}, 1, 4, 3, 1][2,3,1?,1,4,3,1]?and the first card with color?t_3 = 1t3?=1?has position?33;
4. the deck is?[\underline{1}, 2, 3, 1, 4, 3, 1][1?,2,3,1,4,3,1]?and the first card with color?t_4 = 1t4?=1?has position?11;
5. the deck is?[1, 2, 3, 1, \underline{4}, 3, 1][1,2,3,1,4?,3,1]?and the first card with color?t_5 = 4t5?=4?has position?55.

Sponsor

//这题虽然给了两秒，但数据很大，而且一看就必须得嵌套循环
//普通位置对应颜色的筛查应该是行不通，但这题良心的只给了
//50种颜色...这不很明显嘛，用颜色标记位置！走起~
#include <iostream>
#include<algorithm>
using namespace std;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int color[51] = { 0 };
    int n, q;
    cin >> n >> q;
    int eveco, * answer;//储存答案
    answer = (int*)malloc(q * sizeof(int));
    for (int i = 1; i <= n; i++) {//储存颜色种类对应第一次出现的位置
        cin >> eveco;
        if (color[eveco] == 0) {
            color[eveco] = i;
        }
    }
    int t, j = 0;
    for (int i = 0; i < q; i++) {
        cin >> t;
        answer[j++] = color[t];//储存答案
        for (int k = 1; k <= 50; k++) {
            if (color[k] < color[t] && color[k] != 0) {//比筛查样例小的位置+1
                color[k]++;
            }
            else continue;
        }
        color[t] = 1;//把筛查样例放在最前面
    }
    for (int h = 0; h < j; h++) {
        cout << answer[h] << ' ';
    }
    return 0;
}