Codeforces Round #659 (Div. 2) A. Common Prefixes（思维）

The length of the longest common prefix of two strings \(s=s_1 s_2…s_n\) and \(t=t_1t_2…t_m\) is defined as the maximum integer k0≤k≤min(n,m)) such that \(s=s_1 s_2…s_n\) equals to \(t=t_1t_2…t_m\)

Koa the Koala initially has n+1 strings \(s=s_1 s_2…s_{n+1}\)

For each i (1≤i≤n) she calculated \(a_i\)— the length of the longest common prefix of \(s_i\) and \(s_{i+1}\) .

Several days later Koa found these numbers, but she couldn't remember the strings.

So Koa would like to find some strings \(s_1, s_2,…s_{n+1}\)which would have generated numbers \(a_1, a_2,…a_n\). Can you help her?

If there are many answers print any. We can show that answer always exists for the given constraints.

Input

Output

Example

Input

Copy

4
4
1 2 4 2
2
5 3
3
1 3 1
3
0 0 0

Output

Copy

aeren
ari
arousal
around
ari
monogon
monogamy
monthly
kevinvu
kuroni
kurioni
korone
anton
loves
adhoc
problems

首先随便构造一个字符串作为第一个，然后对于每个\(a_i\)，把上一个字符串下标为\(a_i\)的字母换掉就得到当前字符串了（可以直接ASCII码减去'a'再加1，再对26取模）。

#include <bits/stdc++.h>
using namespace std;
int main()
{
	int t;
	cin >> t;
	while(t--)
	{
		int n;
		cin >> n;	
		string s = "";
		for(int i = 0; i < 200; i++) s += 'a';
		cout << s << endl;
		for(int i = 1; i <= n; i++)
		{
			int temp;
			cin >> temp;
			s[temp] = (s[temp] - 'a' + 1) % 26 + 'a';
			cout << s << endl;
		}
	}
	return 0;
}