Problem:
Score : 300 points

Problem Statement

You are given a string s.
Among the different substrings of s, print the K-th lexicographically smallest one.

A substring of s is a string obtained by taking out a non-empty contiguous part in s.
For example, if s = ababc, a, bab and ababc are substrings of s, while ac, z and an empty string are not.
Also, we say that substrings are different when they are different as strings.

Let X = x_{1}x_{2}...x_{n} and Y = y_{1}y_{2}...y_{m} be two distinct strings. X is lexicographically larger than Y if and only if Y is a prefix of X or x_{j} > y_{j} where j is the smallest integer such that x_{j} \neq y_{j}.

Constraints

1 ≤ |s| ≤ 5000

s consists of lowercase English letters.

1 ≤ K ≤ 5

s has at least K different substrings.

Partial Score

200 points will be awarded as a partial score for passing the test set satisfying |s| ≤ 50.

Input

Input is given from Standard Input in the following format:

s
K

Output

Print the K-th lexicographically smallest substring of K.

Sample Input 1

aba
4

Sample Output 1

b

s has five substrings: a, b, ab, ba and aba.
Among them, we should print the fourth smallest one, b.
Note that we do not count a twice.

Sample Input 2

atcoderandatcodeer
5

Sample Output 2

andat

Sample Input 3

z
1

Sample Output 3

z

Solution in C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
#include<string.h>

int main(void){
    char s[5001];
    char *ps = s;
    int k;
    scanf("%s", s);
    scanf("%d", &k);

    char tmp[5000][6];
    char now[6];
    // char ans[5][6];
    char ans[24990][6];
    char buffer[6];
    int index = -1;
    int ls = strlen(s);

    for(int h = 1; h <= 5; h++){
        // h文字分ずつtmpの配列に移す
        for(int i = 0; i < ls-h+1; i++){
            strncpy(tmp[i], ps, h);
            tmp[i][h] = '\0';
            ps++;
        }
        // printf("%d文字ずつtmpに移した結果\n", h);
        // for(int i = 0; i < ls-h+1; i++){
        //     printf("tmp[%d]:%s\n", i, tmp[i]);
        // }
        // 移した後辞書順に昇順にソートする
        for(int i = 0; i < (ls-h+1)-1; i++){
            for(int j = i+1; j < (ls-h+1); j++){
                if(strcmp(tmp[i], tmp[j]) > 0){
                    strcpy(buffer, tmp[i]);
                    strcpy(tmp[i], tmp[j]);
                    strcpy(tmp[j], buffer);
                }
            }
        }
        // printf("%d文字ずつtmpに移した後昇順ソートした結果\n", h);
        // for(int i = 0; i < ls-h+1; i++){
        //     printf("tmp[%d]:%s\n", i, tmp[i]);
        // }
        //重複していない文字列だけansに追加する
        //例えば、bよりもatcoderの方が辞書順でいえば上
        // というケースがもれてる
        //今は文字の長さが短い順でかつ辞書順となっているけど
        //辞書順だけでいいので
        //まず全部ansの配列に入れてから(何個とかは無視)
        //昇順にソートして、kばんめを出す
        for(int i = 0; i < ls-h+1; i++){
            if(i == 0 || (i!=0 && strcmp(now, tmp[i])!=0)){
                // printf("通った!\n");
                // index++;
                // if(index < k){
                //     strcpy(now, tmp[i]);
                //     strcpy(ans[index], now);
                //     // printf("ans[%d]:%s\n", index, ans[index]);
                // }else{
                //     goto END;
                // }
                index+++
                strcpy(now, tmp[i]);
                strcpy(ans[index], now);
            }
        }
            
        ps = s;
    }

    //ここまでですべてansに追加できたので、
    //この後に昇順にソートする
    for(int i = 0; i < index; i++){
        for(int j = i+1; j < index+1; j++){
            if(strcmp(ans[i], ans[j]) > 0){
                strcpy(buffer, ans[i]);
                strcpy(ans[i], ans[j]);
                strcpy(ans[j], buffer);
            }
        }
    }

    printf("%s\n", ans[k-1]);
    return 0;  
}