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 = 2 * 1e5 + 1, INF = 1e9;
int main() {
  int n, k;
  char c[MAX], c1;
  int niz[MAX], poz, br;
  string s, s1;
  bitset<27> b;
  vector<int> v;
  cin >> n >> k;
  cin >> s1;
  for (int i = 0; i <= n - 1; i++) {
    niz[i + 1] = s1[i] - 'a' + 1;
    b[s1[i] - 'a' + 1] = 1;
  }
  for (int i = 1; i <= 26; i++)
    if (b[i]) v.push_back(i);
  if (n >= k) {
    br = k;
    while ((niz[br] == v.back()) && (br > 0)) {
      br--;
    }
    for (int i = 1; i <= br - 1; i++) s.push_back(niz[i] + 'a' - 1);
    poz = 0;
    while (v[poz] != niz[br]) poz++;
    s.push_back(v[poz + 1] + 'a' - 1);
    for (int i = br + 1; i <= k; i++) s.push_back(v[0] + 'a' - 1);
    cout << s;
  } else {
    for (int i = 0; i <= n - 1; i++) s.push_back(niz[i + 1] + 'a' - 1);
    for (int i = n; i <= k - 1; i++) s.push_back(v[0] + 'a' - 1);
    cout << s;
  }
  return 0;
}
