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 a[200020];
map<char, int> ma;
int main() {
  int n, m, mrk;
  cin >> n >> m;
  char p, mn = 'z', ch;
  mrk = n;
  for (int i = 1; i <= n; i++)
    cin >> ch, a[i] = ch, ma[ch] = 1, mn = min(mn, a[i]);
  if (m > n) {
    for (int i = 1; i <= n; i++) cout << a[i];
    for (int i = n + 1; i <= m; i++) cout << mn;
    return 0;
  }
  for (int i = m; i >= 1; i--) {
    char c = a[i];
    bool k = 0;
    for (char j = c; j <= 'z'; j++)
      if (j != c && ma[j]) {
        mrk = i;
        p = j, k = 1;
        break;
      }
    if (k) break;
  }
  for (int i = 1; i <= mrk - 1; i++) cout << a[i];
  cout << p;
  for (int i = mrk + 1; i <= m; i++) cout << mn;
}
