B. Prinzessin der Verurteilung

Problem - 1536B - Codeforces

I, Fischl, Prinzessin der Verurteilung, descend upon this land by the call of fate an — Oh, you are also a traveler from another world? Very well, I grant you permission to travel with me.

It is no surprise Fischl speaks with a strange choice of words. However, this time, not even Oz, her raven friend, can interpret her expressions! Maybe you can help us understand what this young princess is saying?

You are given a string of nn lowercase Latin letters, the word that Fischl just spoke. You think that the MEX of this string may help you find the meaning behind this message. The MEX of the string is defined as the shortest string that doesn't appear as a contiguous substring in the input. If multiple strings exist, the lexicographically smallest one is considered the MEX. Note that the empty substring does NOT count as a valid MEX.

A string aa is lexicographically smaller than a string bb if and only if one of the following holds:

• aa is a prefix of bb, but a≠ba≠b;
• in the first position where aa and bb differ, the string aa has a letter that appears earlier in the alphabet than the corresponding letter in bb.

A string aa is a substring of a string bb if aa can be obtained from bb by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.

Find out what the MEX of the string is!

Input

Each test contains multiple test cases. The first line contains the number of test cases tt (1≤t≤10001≤t≤1000). Description of the test cases follows.

The first line of each test case contains an integer nn (1≤n≤10001≤n≤1000) — the length of the word. The second line for each test case contains a single string of nn lowercase Latin letters.

The sum of nn over all test cases will not exceed 10001000.

Output

For each test case, output the MEX of the string on a new line.

Example

input

Copy

3
28
qaabzwsxedcrfvtgbyhnujmiklop
13
cleanairactbd
10
aannttoonn

output

Copy

ac
f
b

 mex 为字符串中未出现的最短的字符串

直接暴力

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
// ctrl+shift+C 注释
//ctrl+shift+x 取消
//#define int long long
#define YES cout<<"YES"<
#define NO cout<<"NO"<
#define fast ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//int k = min_element(a + 1, a + 1 + n) - a
typedef long long ll;
typedef pair<int,int> PII;
const int N=2e5+10;
const ll M=1e18+10;
const int mod=1e9+7;
int a[N],sum[N];
priority_queue<int,vector<int>,greater<int> >pq;
set<int>se;
map<char,int>mp;
queue<int>qu;
vector<int>v;
deque<int>de;
//struct Range
//{
//    int l,r;
//    bool operator< (const Range &w)const
//    {
//        return l
//    }
//}range[N];
//-----------------------------分割线---------------------------------------------------
int n;
string s;
void solve()
{
    cin>>n>>s;
    mp.clear();
    int k=0;
  for(int i=0;i
  {
      if(!mp[s[i]])k++;
      mp[s[i]]++;
  }
  if(k<26)//没出现的
  {
      for(int i=0;i<26;i++)
      {
          if(!mp['a'+i])
          {
              printf("%c\n",'a'+i);
              return;
          }
      }
  }
  string ss="aa";//二个的
  for(int i=0;i<26;i++)
  {
      ss[0]='a'+i;
      for(int j=0;j<26;j++)
      {
         ss[1]='a'+j;
          if(s.find(ss)==-1)
             {
                 cout<
                    return;
             }
      }
  }
  ss="aaa";//三个的
   for(int i=0;i<26;i++)
  {
      ss[0]='a'+i;
      for(int j=0;j<26;j++)
      {
         ss[1]='a'+j;
         for(int m=0;m<26;m++)
          {
              ss[2]='a'+m;
              if(s.find(ss)==-1)
             {
                 cout<
                    return;
             }
        }
      }
  }
}
int main()
{
   int t=1;
   cin>>t;
   while(t--)
   {
       solve();
   }
}