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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,sse3,sse4,popcnt,abm,mmx")
using namespace std;
const double eps = 0.000001;
const long double pi = acos(-1);
const int maxn = 1e7 + 9;
const int mod = 1e9 + 7;
const long long MOD = 1e18 + 9;
const long long INF = 1e18 + 123;
const int inf = 2e9 + 11;
const int mxn = 1e6 + 9;
const int N = 3e5 + 123;
const int PRI = 555557;
const int pri = 997;
int tests = 1;
int n, m, k;
string s;
int cnt[31];
int cur, res;
int ans;
set<int> q;
inline void Solve() {
  cin >> n >> k;
  cin >> s;
  for (int i = 0; i < s.size(); i++) cnt[s[i] - 'a'] = 1;
  if (k > n) {
    cout << s;
    for (int j = n + 1; j <= k; j++)
      for (int i = 0; i < 26; i++) {
        if (cnt[i]) {
          cout << char(i + 'a');
          break;
        }
      }
    exit(0);
  }
  n = k;
  for (int i = n - 1; i >= 0; i--) {
    for (int j = s[i] - 'a' + 1; j < 26; j++) {
      if (cnt[j]) {
        s[i] = j + 'a';
        cur = i;
        break;
      }
    }
    if (cur) break;
  }
  for (int i = 0; i <= cur; i++) cout << s[i];
  cur++;
  for (; cur < n; cur++) {
    for (int i = 0; i < 26; i++) {
      if (cnt[i]) {
        cout << char(i + 'a');
        break;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  while (tests--) {
    Solve();
  }
  return 0;
}
