1206_A. Choose Two Numbers
You are given an array A, consisting of n positive integers a_1, a_2, ..., a_n, and an array B, consisting of m positive integers b_1, b_2, ..., b_m. 

Choose some element a of A and some element b of B such that a+b doesn't belong to A and doesn't belong to B. 

For example, if A = [2, 1, 7] and B = [1, 3, 4], we can choose 1 from A and 4 from B, as number 5 = 1 + 4 doesn't belong to A and doesn't belong to B. However, we can't choose 2 from A and 1 from B, as 3 = 2 + 1 belongs to B.

It can be shown that such a pair exists. If there are multiple answers, print any.

Choose and print any such two numbers.

Input

The first line contains one integer n (1≤ n ≤ 100) — the number of elements of A.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 200) — the elements of A.

The third line contains one integer m (1≤ m ≤ 100) — the number of elements of B.

The fourth line contains m different integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ 200) — the elements of B.

It can be shown that the answer always exists.

Output

Output two numbers a and b such that a belongs to A, b belongs to B, but a+b doesn't belong to nor A neither B.

If there are multiple answers, print any.

Examples

Input


1
20
2
10 20


Output


20 20

Input


3
3 2 2
5
1 5 7 7 9


Output


3 1


Input


4
1 3 5 7
4
7 5 3 1


Output


1 1

Note

In the first example, we can choose 20 from array [20] and 20 from array [10, 20]. Number 40 = 20 + 20 doesn't belong to any of those arrays. However, it is possible to choose 10 from the second array too.

In the second example, we can choose 3 from array [3, 2, 2] and 1 from array [1, 5, 7, 7, 9]. Number 4 = 3 + 1 doesn't belong to any of those arrays.

In the third example, we can choose 1 from array [1, 3, 5, 7] and 1 from array [7, 5, 3, 1]. Number 2 = 1 + 1 doesn't belong to any of those arrays.
```python
n=int(input())
a=list(map(int,input().split()))
m=int(input())
b=list(map(int,input().split()))
t=[];o=[]
for i in a:
    for j in b:
        t.append(i+j)
        o.append((i,j))

for i in range(len(t)):
    if (t[i] not in a) and (t[i] not in b):
        print(o[t.index(t[i])][0])
        print(o[t.index(t[i])][1])
        break;
        

```