Eva is trying to make her own color stripe out of a given one. She would like to keep only her favorite colors in her favorite order by cutting off those unwanted pieces and sewing the remaining parts together to form her favorite color stripe.

It is said that a normal human eye can distinguish about less than 200 different colors, so Eva's favorite colors are limited. However the original stripe could be very long, and Eva would like to have the remaining favorite stripe with the maximum length. So she needs your help to find her the best result.

Note that the solution might not be unique, but you only have to tell her the maximum length. For example, given a stripe of colors {2 2 4 1 5 5 6 3 1 1 5 6}. If Eva's favorite colors are given in her favorite order as {2 3 1 5 6}, then she has 4 possible best solutions {2 2 1 1 1 5 6}, {2 2 1 5 5 5 6}, {2 2 1 5 5 6 6}, and {2 2 3 1 1 5 6}.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integerN(≤200) which is the total number of colors involved (and hence the colors are numbered from 1 toN). Then the next line starts with a positive integerM(≤200) followed byMEva's favorite color numbers given in her favorite order. Finally the third line starts with a positive integerL(≤10⁴

Output Specification:

For each test case, simply print in a line the maximum length of Eva's favorite stripe.

Sample Input:

6
5 2 3 1 5 6
12 2 2 4 1 5 5 6 3 1 1 5 6

Sample Output:

7

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        String s = reader.readLine();
        int N = Integer.parseInt(s);        //N種顏色
        String s1 = reader.readLine();
        String[] s2 = s1.split(" ");
        int M = Integer.parseInt(s2[0]);
        int[] favorite = new int[M];       //最喜歡的顏色序列
        for (int i = 0; i < M; i++) {
            favorite[i] = Integer.parseInt(s2[i + 1]);
        }

        String s3 = reader.readLine();
        String[] s4 = s3.split(" ");
        int L = Integer.parseInt(s4[0]);
        int[] gaven = new int[L];         //給定條紋
        for (int i = 0; i < L; i++) {
            gaven[i] = Integer.parseInt(s4[i + 1]);
        }

        System.out.println(maxLongCommonSubsquence(favorite, gaven, M, L));
    }

    public static int maxLongCommonSubsquence(int[] arr1, int[] arr2, int n1, int n2){
        int[][] dp = new int[n1 + 1][n2 + 1];

        for (int i = 1; i <= n1; i++) {
            for (int j = 1; j <= n2; j++){
                int max = Math.max(dp[i - 1][j], dp[i][j - 1]);
                if (arr1[i-1] == arr2[j-1]){
                    dp[i][j] = max + 1;
                }else {
                    dp[i][j] = max;
                }
            }
        }
        return dp[n1][n2];
    }
}