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;
char s[100050], t[100050];
int vis[30], cnt, mp[30];
int n, k;
char maxn;
int pos;
int main() {
  cin >> n >> k;
  scanf("%s", s + 1);
  for (int i = 1; i <= n; ++i) {
    if (!mp[s[i] - 'a']) {
      mp[s[i] - 'a'] = 1;
      vis[++cnt] = s[i] - 'a';
    }
  }
  sort(vis + 1, vis + cnt + 1);
  if (k > n) {
    for (int i = 1; i <= n; ++i) t[i] = s[i];
    for (int i = n + 1; i <= k; ++i) {
      t[i] = vis[1] + 'a';
    }
    printf("%s", t + 1);
    return 0;
  }
  if (k <= n) {
    bool flag = false, flag2 = false;
    if (s[k] == vis[cnt] + 'a') {
      flag = true;
      t[k] = vis[1] + 'a';
    } else {
      int pp;
      for (int j = 1; j <= cnt; ++j) {
        if (vis[j] + 'a' > s[k]) {
          pp = j;
          break;
        }
      }
      t[k] = vis[pp] + 'a';
    }
    for (int i = k - 1; i; --i) {
      if (s[i] != vis[cnt] + 'a' && flag) {
        flag = false;
        flag2 = true;
        int pp;
        for (int j = 1; j <= cnt; ++j) {
          if (vis[j] + 'a' > s[i]) {
            pp = j;
            break;
          }
        }
        t[i] = vis[pp] + 'a';
      } else if (s[i] == vis[cnt] + 'a' && flag) {
        t[i] = vis[1] + 'a';
      } else if (!flag) {
        flag2 = true;
        t[i] = s[i];
      }
    }
    printf("%s", t + 1);
    return 0;
  }
  return 0;
}
