import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeSet;

// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        // 注意 hasNext 和 hasNextLine 的区别
        while (in.hasNextInt()) { // 注意 while 处理多个 case
            int n1 = in.nextInt();
            int[]arr1 = new int[n1];
            for (int i = 0; i < n1; i++) {
                arr1[i] = in.nextInt();
            }
            int n2 = in.nextInt();
            Set<Integer>set = new TreeSet<>();
            for (int i = 0; i < n2; i++) {
                set.add(in.nextInt());
            }
		  //转化成list集合方便遍历
            List<Integer>list = new ArrayList<>(set);
		  //用于记录次数
            List<Integer>counts = new ArrayList<>();
		  //用于记录元素下标和元素
            List<Integer>datas = new ArrayList<>();

            for (int i = 0; i < list.size(); i++) {
                String s2 = list.get(i) + "";
                int count = 0;
                for (int j = 0; j < arr1.length; j++) {
                    String s1 = arr1[j] + "";
                    if (s1.contains(s2)) {
                        count++;
                        datas.add(j);
                        datas.add(arr1[j]);
                    }
                }
                counts.add(count );
            }
            int sum = 0;
            for (Integer count : counts) {
			  //这里要做一个大于0的判断,否则次数会出错
                if(count>0)
                sum = sum+count*2+2 ;
            }
            System.out.print(sum + " ");
            int j = 0;
            for (int i = 0; i < counts.size(); i++) {
			  //这道题的难点在于输入输出,分段遍历集合时上限是每次遍历长度的累加值,否则就会遍历不到数据
                int len =j+counts.get(i)*2;
                if (counts.get(i) > 0) {
                    System.out.print(list.get(i) + " ");
                    System.out.print(counts.get(i) + " ");
                    for (; j < len; j++) {
                        System.out.print(datas.get(j) + " ");
                    }
                }
            }
        }
    }
}