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 bool debug = false;
const bool INJUDGE = false;
set<int> f;
int l, l2;
string s;
string s2;
bool ismax[100005];
int maxc, minc;
int nxt[256];
void _fill() {
  for (int i = s2.size() + 1; i <= l2; i++) s2 += (char)minc;
  cout << s2 << endl;
  exit(0);
}
void back_to_find_and_fill() {
  for (int i = s2.size() - 1; i >= 0; i--) {
    if (nxt[s2[i]] != -1) {
      s2[i] = nxt[s2[i]];
      s2 = s2.substr(0, i + 1);
      _fill();
    }
  }
}
int main() {
  cin >> l >> l2;
  cin >> s;
  minc = 1000;
  for (int i = 0; i < s.size(); i++)
    f.insert(s[i]), maxc = max(maxc, (int)s[i]), minc = min(minc, (int)s[i]);
  for (int i = s.size() - 1; i >= 0; i--) {
    if (s[i] != maxc) break;
    ismax[i] = 1;
  }
  for (int i = 'a'; i <= 'z'; i++) {
    if (i >= maxc) nxt[i] = -1;
    for (int j = i + 1; j <= 'z'; j++) {
      if (f.find(j) != f.end()) {
        nxt[i] = j;
        break;
      }
    }
  }
  if (l == l2) {
    for (int i = 0; i < l2; i++) {
      if (!ismax[i])
        s2 += s[i];
      else
        back_to_find_and_fill();
    }
    back_to_find_and_fill();
  }
  if (l2 > l) {
    s2 = s;
    _fill();
  }
  if (l2 < l) {
    for (int i = 0; i < l2; i++) {
      if (!ismax[i])
        s2 += s[i];
      else
        back_to_find_and_fill();
    }
    back_to_find_and_fill();
  }
  return 0;
}
