Codeforces Round #466 (Div. 2) Phone Numbers

C. Phone Numbers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

And where the are the phone numbers?

You are given a string s consisting of lowercase English letters and an integer k. Find the lexicographically smallest string t of length k, such that its set of letters is a subset of the set of letters of s and s is lexicographically smaller than t.

It's guaranteed that the answer exists.

Note that the set of letters is a set, not a multiset. For example, the set of letters of abadaba is {a, b, d}.

String p is lexicographically smaller than string q, if p is a prefix of q, is not equal to q or there exists i, such that pi < qi and for all j < i it is satisfied that pj = qj. For example, abc is lexicographically smaller than abcd , abd is lexicographically smaller than abec, afa is not lexicographically smaller than ab and a is not lexicographically smaller than a.

Input

The first line of input contains two space separated integers n and k (1 ≤ n, k ≤ 100 000) — the length of s and the required length of t.

The second line of input contains the string s consisting of n lowercase English letters.

Output

Output the string t conforming to the requirements above.

It's guaranteed that the answer exists.

Examples

input

Copy

3 3
abc

output

aca

input

Copy

3 2
abc

output

ac

input

Copy

3 3
ayy

output

yaa

input

Copy

2 3
ba

output

baa

Note

In the first example the list of strings t of length 3, such that the set of letters of t is a subset of letters of s is as follows: aaa, aab, aac, aba, abb, abc, aca, acb, .... Among them, those are lexicographically greater than abc: aca, acb, .... Out of those the lexicographically smallest is aca.

 

找规律, 构造题

 构造方法

k>n

　　贪心的加最小的，前面和s一样

k<=n

　　 从后往前扫描

　　如果是'z'

　　则用最小的

　　不然用可行的比原字母大一的

　　并结束构造

　　让前面的和s相同

代码:

#include<iostream>
using namespace std;
#include<cstdio>
#include<cstring>
#include<map>
map<int,int> maps;
char s[300000];
char a[300000];
int main(){
    int n,k;
    scanf("%d%d",&n,&k);
    scanf("%s",a);
    char MIN = 'a';
    for(int i=0;i<n;i++){
        if(i==0||a[i]<MIN)
            MIN = a[i];
        s[i]=a[i];
        maps[(int)a[i]]=1;
    }
    for(int i=n;i<k;i++){
        s[i]=MIN;
    }
    int pt=0;
    if(n>=k){
        for(int i=k-1;i>=0&&pt==0;i--){
            int p=s[i];
            for(int j=p+1;j<='z'+1&&pt==0;j++){
                if(j=='z'+1){
                    s[i]=MIN;
                }
                else if(maps[j]){
                    s[i]=j;
                    pt=1;
                    break;
                }
            }
        }
    }
    for(int i=0;i<k;i++)
        putchar(s[i]);
    return 0;
}