Happy new year! The year 2020 is also known as Year Gyeongja (경자년, gyeongja-nyeon) in Korea. Where did the name come from? Let's briefly look at the Gapja system, which is traditionally used in Korea to name the years.

There are two sequences of n strings s_1, s_2, s_3, …, s_{n} and m strings t_1, t_2, t_3, …, t_{m}. These strings contain only lowercase letters. There might be duplicates among these strings.

Let's call a concatenation of strings x and y as the string that is obtained by writing down strings x and y one right after another without changing the order. For example, the concatenation of the strings "code" and "forces" is the string "codeforces".

The year 1 has a name which is the concatenation of the two strings s_1 and t_1. When the year increases by one, we concatenate the next two strings in order from each of the respective sequences. If the string that is currently being used is at the end of its sequence, we go back to the first string in that sequence.

For example, if n = 3, m = 4, s = {"a", "b", "c"}, t = {"d", "e", "f", "g"}, the following table denotes the resulting year names. Note that the names of the years may repeat.

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You are given two sequences of strings of size n and m and also q queries. For each query, you will be given the current year. Could you find the name corresponding to the given year, according to the Gapja system?

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 20).

The next line contains n strings s_1, s_2, …, s_{n}. Each string contains only lowercase letters, and they are separated by spaces. The length of each string is at least 1 and at most 10.

The next line contains m strings t_1, t_2, …, t_{m}. Each string contains only lowercase letters, and they are separated by spaces. The length of each string is at least 1 and at most 10.

Among the given n + m strings may be duplicates (that is, they are not necessarily all different).

The next line contains a single integer q (1 ≤ q ≤ 2 020).

In the next q lines, an integer y (1 ≤ y ≤ 10^9) is given, denoting the year we want to know the name for.

Output

Print q lines. For each line, print the name of the year as per the rule described above.

Example

Input


10 12
sin im gye gap eul byeong jeong mu gi gyeong
yu sul hae ja chuk in myo jin sa o mi sin
14
1
2
3
4
10
11
12
13
73
2016
2017
2018
2019
2020


Output


sinyu
imsul
gyehae
gapja
gyeongo
sinmi
imsin
gyeyu
gyeyu
byeongsin
jeongyu
musul
gihae
gyeongja

Note

The first example denotes the actual names used in the Gapja system. These strings usually are either a number or the name of some animal.
// Java program For handling Input/Output 
import java.io.*; 
import java.util.*;
import java.math.*;
//import java.awt.*;

public class Main 
{ 
    Scanner sc;
    Map<Integer, List<Integer>> graph = new HashMap<>();
    //actual logic
    void solve(){
        int n = ni();
        int m = ni();

        String s[] = new String[n];
        String t[] = new String[m];
        ns();
        for(int i = 0;i < n;i++)
            s[i] = sc.next();
        for(int i = 0;i < m;i++)
            t[i] = sc.next();

        int q = ni();
        while(--q >= 0)
        {
            int target = ni()-1;
            System.out.println(s[target%n]+t[target%m]);
        }
    }

    //constructor
    Main(){
        try{  
            sc = new Scanner(System.in);
        }catch(Exception e){
            System.out.println(e);
        }
    }


    //main metohd
    public static void main(String[] args) 
    { 
     new Main().solve();
 }


    //Utility methods
 void graphInput(int n)
 {
    for(int i = 0;i < n;i++)
    {
        int a = ni();
        int b = ni();

        if(!graph.containsKey(a))
        {
            List<Integer> l = new ArrayList<>();
            l.add(b);
            graph.put(a, l);
        }
        else
            graph.get(a).add(b);
    }
    System.out.println(graph);
}

void bfs(int nodes)
{
    Queue<Integer> q = new LinkedList<>();
    int visited[] = new int[nodes+1];

    q.add(1);
    while(!q.isEmpty())
    {
        System.out.print(q.peek() + " ");
        visited[q.peek()] = 1;
        if(graph.get(q.peek()) != null)
        {
            for(int i: graph.get(q.peek()))
            {
                if(visited[i] != 1)
                    q.add(i);
            }
        }

        q.poll();
    }
}
        //input an integer from scanner
int ni()
{
    int a = sc.nextInt();
    return a;
}
        //input a long from scanner
long nl()
{
    return sc.nextLong();
}
        //input a float from scanner
float nf(){
    float a = sc.nextFloat();
    return a;
}
        //input a double from scanner
double nd(){
    double a = sc.nextDouble();
    return a;
}
        //input a sentence from scanner
String ns(){
    return sc.nextLine();
}
        //converts a string to stringtokenizer
StringTokenizer nst(String s){
    return new StringTokenizer(s);
}
        //input an intger array
void ia(int a[]){
    for(int i = 0;i < a.length;i++)
        a[i] = sc.nextInt();
}
        //
void la(long a[]){
    for(int i = 0;i < a.length;i++)
        a[i] = sc.nextLong();
}
        //input a float array
void fa(float a[]){
    for(int i = 0;i < a.length;i++)
        a[i] = sc.nextFloat();
}
        //input a double array
void da(double a[]){
    for(int i = 0;i < a.length;i++)
        a[i] = sc.nextDouble();
}
        //input a string array either with all empty input or from scanner
void sa(String a[], boolean empty){
    if(empty)
    {
        for(int i = 0;i < a.length;i++)
        {
            a[i] = "";
        }
    }
    else
    {
        for(int i = 0;i < a.length;i++)
        {
            a[i] = ns();
        }
    }
}
        //input two dimensional integer array
void ida(int a[][], int n, int m)
{
    for(int i = 0;i < n;i++)
    {
        for(int j = 0;j < m;j++)
        {
            a[i][j] = ni();
        }
    }
}
        //input two dimentional long array
void lda(long a[][], int n, int m)
{
    for(int i = 0;i < n;i++)
    {
        for(int j = 0;j < m;j++)
        {
            a[i][j] = nl();
        }
    }
}
        //input two dimensional double array
void dda(double a[][], int n, int m)
{
    for(int i = 0;i < n;i++)
    {
        for(int j = 0;j < m;j++)
        {
            a[i][j] = nd();
        }
    }
}
        //convert string to integer
int stoi(String s){
    return Integer.parseInt(s);
}
        //convert string to double
double stod(String s){
    return Double.parseDouble(s);
}
        //find minimum in a long array
long lmin(long a[])
{
    long min = Long.MAX_VALUE;
    for(int i = 0;i < a.length;i++)
    {
        if(min > a[i])
            min = a[i];
    }
    return min;
}
        //find minimum in a integer array
int imin(int a[])
{
    int min = Integer.MAX_VALUE;
    for(int i = 0;i < a.length;i++)
    {
        if(min > a[i])
            min = a[i];
    }
    return min;
}
        //find maximum in a long array
long lmax(long a[])
{
    long max = Long.MIN_VALUE;
    for(int i = 0;i < a.length;i++)
    {
        if(max < a[i])
            max = a[i];
    }
    return max;
}
        //find maximum in an integer array
int imax(int a[])
{
    int max = Integer.MIN_VALUE;
    for(int i = 0;i < a.length;i++)
    {
        if(max < a[i])
            max = a[i];
    }
    return max;
}
        //check whether an element is present in an integer array by binary search
boolean ibs(int a[], int toFind)
{
    Arrays.sort(a);
    int min = 0;
    int max = a.length-1;
    boolean found = false;

    while(min <= max && !found)
    {
        int mid = min + (max-min)/2;
        if(a[mid] == toFind)
        {
           found = true;
       }
       else if(toFind > a[mid])
       {
        min = mid+1;
    }
    else
    {
        max = mid-1;
    }

}
return found;
}
        //check whether an element is present in a long array
boolean lbs(long a[], long toFind)
{
    Arrays.sort(a);
    int min = 0;
    int max = a.length-1;
    boolean found = false;

    while(min <= max && !found)
    {
        int mid = min + (max-min)/2;
        if(a[mid] == toFind)
        {
           found = true;
       }
       else if(toFind > a[mid])
       {
        min = mid+1;
    }
    else
    {
        max = mid-1;
    }

}
return found;
}

int stb(String s)
{
    int sum = 0;
    int k = 0;
    for(int i = s.length()-1;i >= 0;i--)
    {
        sum += stoi(s.charAt(i)+"") * Math.pow(2, k++);
    }
    return sum;
}

}