Codeforces 940c【贪心】

  • 2018-03-06
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C. Phone Numbers

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 abecafa 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
 3 3
 abc
output
 aca
input
 3 2
 abc
output
 ac
input
 3 3
 ayy
output
 yaa
input
 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: aaaaabaacabaabbabcacaacb. Among them, those are lexicographically greater than abcacaacb. Out of those the lexicographically smallest is aca.


Code

#include <bits/stdc++.h>
using namespace std;
set<char> st;

int main()
{
    int n, k;
    char s[100005], t[100005];
    scanf("%d%d", &n, &k);
    getchar();
    scanf("%s", s);
    int len = strlen(s);
    for (int i = 0; i < len; i++)
        st.insert(s[i]);
    int tot = st.size();
    set<char>::iterator it, it2;
    if (n < k)
    {
        printf("%s", s);
        char ch = *st.begin();
        for (int i = 0; i < k - n; i++)
            printf("%c", ch);
        printf("\n");
    }else
    {
        for (int i = len - 1 - (n - k); i >= 0; i--)
        {
            it = st.find(s[i]);
            it++;
            if (it == st.end())
            {
                s[i] = *st.begin();
            }else
            {
                s[i] = *it;
                break;
            }
        }
        for (int i = 0; i < k; i++)
            printf("%c", s[i]);
        printf("\n");
    }

    return 0;
}

Origin

Codeforces 940c

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