Codeforces Round #440 div2 A. Search for Pretty Integers

A. Search for Pretty Integers

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two lists of non-zero digits.

Let's call an integer pretty if its (base 10) representation has at least one digit from the first list and at least one digit from the second list. What is the smallest positive pretty integer?

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 9) — the lengths of the first and the second lists, respectively.

The second line contains n distinct digits a1, a2, ..., an (1 ≤ ai ≤ 9) — the elements of the first list.

The third line contains m distinct digits b1, b2, ..., bm (1 ≤ bi ≤ 9) — the elements of the second list.

Output

Print the smallest pretty integer.

Examples

input

2 3
4 2
5 7 6

output

25

input

8 8
1 2 3 4 5 6 7 8
8 7 6 5 4 3 2 1

output

1

Note

In the first example 25, 46, 24567 are pretty, as well as many other integers. The smallest among them is 25. 42 and 24 are not pretty because they don't have digits from the second list.

In the second example all integers that have at least one digit different from 9 are pretty. It's obvious that the smallest among them is 1, because it's the smallest positive integer.

题意 输入n，m，接下来两行为两个序列，如果两个序列中有相同的输出相同的最小的那个数，如果没有就从两个序列中分别取出一个数，输出这两个数组成的最小两位数。

桶排序就好

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <cmath>
using namespace std;

#define ll long long 
#define mem(a, b) memset(a, b, sizeof(a))
#define inf 0x3f3f3f3f
#define mod 10000000007
#define Maxn 10000005
#define debug() puts("F*** you every!");

int main()
{
	int n, m, a[50], b[50], w;
	mem(a, 0);
	mem(b, 0);
	scanf("%d%d", &n, &m);
	for (int i = 0; i < n; i ++) {
		scanf("%d", &w);
		a[w]++;
	}
	for (int i = 0; i < m; i++) {
		scanf("%d", &w);
		b[w]++;
	}
	int flag = 0;
	for (int i = 1; i < 10; i ++) {
		if (a[i] != 0 && b[i] != 0)
		{
			printf("%d\n", i);
			flag = 1;
			break;
		}
	}
	int s1, s2;
	if (flag == 0) {
		for (int i = 0; i < 10; i ++) { 
			if (a[i] != 0) {
				s1 = i;break;
			}
		}
		for (int i = 0; i < 10; i ++) { 
			if (b[i] != 0) {
				s2 = i;break;
			}
		}
		printf("%d%d\n", s1 > s2 ? s2 : s1, s1 <= s2 ? s2 : s1);
	}
		
	
}