Little Daniel loves to play with strings! He always finds different ways to have fun with strings! Knowing that, his friend Kinan decided to test his skills so he gave him a string S and asked him Q questions of the form:

If all distinct substrings of string S were sorted lexicographically, which one will be the K-th smallest?

After knowing the huge number of questions Kinan will ask, Daniel figured out that he can‘t do this alone. Daniel, of course, knows your exceptional programming skills, so he asked you to write him a program which given S will answer Kinan‘s questions.

Example:

S = "aaa" (without quotes)
substrings of S are "a" , "a" , "a" , "aa" , "aa" , "aaa". The sorted list of substrings will be:

Input

In the first line there is Kinan‘s string S (with length no more than 90000 characters). It contains only small letters of English alphabet. The second line contains a single integer Q (Q <= 500) , the number of questions Daniel will be asked. In the next Q lines a single integer K is given (0 < K < 2^31).

Output

Output consists of Q lines, the i-th contains a string which is the answer to the i-th asked question.

Example

Input:
aaa
2
2
3

Output:
aa
aaa

Edited: Some input file contains garbage at the end. Do not process them.

讓你求按字典序排名第k名的字符串是什麽

dp處理出每個點往下走能夠走出多少個串。
f[i]=sigma(f[ch[i][c])+1
這個可以按Max排序之後倒著推就好了。
詢問的時候看一下走下去個數是否<=k，是的話就走下去，然後–k；否則就找下一條邊。

 1 #include"bits/stdc++.h"
 2 using namespace std;
 3 const int N = 101000;
 4 #define int long long
 5 struct node
 6 {
 7     int fa;
 8     int ch[26];
 9     int len;
10 }dian[N<<1];
11 int cnt[N<<1];
12 int f[N<<1];
13 int last=1;int tot=1;
14 
15 inline void add(int c)
16 {
17   int p=last; int np=last=++tot;
18   dian[np].len=dian[p].len+1;
19   for(;p&&!dian[p].ch[c];p=dian[p].fa)dian[p].ch[c]=np;
20   if(!p) dian[np].fa=1,cnt[1]++;
21   else
22   {
23      int q=dian[p].ch[c];
24      if(dian[q].len == dian[p].len+1)dian[np].fa=q,cnt[q]++;
25      else
26      {
27          int nq=++tot;
28          dian[nq]=dian[q];
29          dian[nq].len=dian[p].len+1;
30          dian[q].fa=dian[np].fa=nq;
31 ;         cnt[nq]+=2;
32          for(;p&&dian[p].ch[c] == q;p=dian[p].fa)dian[p].ch[c]=nq;
33      }
34   }
35 }
36 
37 
38 void dfs(int x)
39 {
40     if(f[x]) return ;
41     f[x]=1;
42     for(int i=0;i<26;i++)
43       {
44            if(dian[x].ch[i])
45            {dfs(dian[x].ch[i]);
46             f[x]+= f[dian[x].ch[i]];
47 
48            }
49       }
50 }
51 void que()
52 {
53     int x;cin>>x; int now=1;
54     while(x)
55     {
56         for(int i=0;i<26;i++)if(dian[now].ch[i])
57         {
58             if(f[dian[now].ch[i]]>=x)
59             {
60                 putchar(‘a‘+i); --x;now=dian[now].ch[i];
61                 break;
62             }
63             else x-=f[dian[now].ch[i]];
64         }
65     }
66     cout<<‘\n‘;
67 }
68 signed main()
69 {
70     //  freopen("datte.txt","r",stdin);
71    //freopen("my.out","w",stdout);
72    string s;cin>>s; int len =s.length();
73     for(int i=0;i<len;i++)add(s[i]-‘a‘);
74    
75     dfs(1);  int m;cin>>m;
76     while(m--)que();
77 }

D - Lexicographical Substring Search