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.
import sys 
# sys.setrecursionlimit(10**6) 
from sys import stdin, stdout
import bisect            #c++ upperbound
import math
import heapq
def modinv(n,p):
    return pow(n,p-2,p)
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
import math  
def Divisors(n) : 
    l = []  
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                l.append(n//i)
    return l
"""*******************************************************"""
def main():
    n,m=cin()
    a=[]
    b=[]
    s=sin()
    ss=sin()
    d=""
    for i in s:
        if(i==" "):
            a.append(d)
            d=""
            continue
        d+=i
    a.append(d)
    d=""
    for i in ss:
        if(i==" "):
            b.append(d)
            d=""
            continue
        d+=i
    b.append(d)
    # print(a,b)
    t=inin()
    for _ in range(t):
        p=inin()
        q=p
        q=q%n
        p=p%m
        print(a[q-1]+b[p-1])
######## Python 2 and 3 footer by Pajenegod and c1729
 
# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.
 
# So on cf, use PyPy2 for best string performance.
 
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
 
import os, sys
from io import IOBase, BytesIO
 
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
 
# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'
 
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'0' [0]:
        A.append(sign*numb)
    return A
 
if __name__== "__main__":
  main()