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

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: 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.
#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 6;
const int MOD = 1e9 + 7;
const long long INF = 1e13 + 77;
int n, k;
char in[MAX];
int freq[333];
vector<int> s;
vector<char> res;
int dp[MAX][3];
int solve(int id, int flg) {
  if (id == k) return (flg == 1);
  int &ret = dp[id][flg];
  if (ret != -1) return ret;
  ret = 0;
  for (int i = 0; i < s.size(); ++i) {
    if (!flg && s[i] < in[id]) continue;
    if (flg)
      ret = ret | solve(id + 1, flg);
    else
      ret = ret | solve(id + 1, s[i] > in[id]);
  }
  return ret;
}
void print(int id, int flg) {
  if (id == k) return;
  int ret = dp[id][flg];
  for (int i = 0; i < s.size(); ++i) {
    if (!flg && s[i] < in[id]) continue;
    if (flg) {
      if (solve(id + 1, flg) == 1) {
        res.push_back(s[i]);
        print(id + 1, flg);
        return;
      }
    } else {
      if (solve(id + 1, s[i] > in[id])) {
        res.push_back(s[i]);
        print(id + 1, s[i] > in[id]);
        return;
      }
    }
  }
}
int main() {
  scanf("%d", &n);
  scanf("%d", &k);
  scanf(" %s", in);
  for (int i = 0; i < n; ++i) {
    if (!freq[in[i]]) {
      s.push_back(in[i]);
      freq[in[i]] = 1;
    }
  }
  for (int i = n; i < k; ++i) in[i] = '0';
  sort(s.begin(), s.end());
  memset(dp, -1, sizeof dp);
  solve(0, 0);
  print(0, 0);
  for (int i = 0; i < res.size(); ++i) putchar(res[i]);
  return 0;
}
