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 long long M = 1e9 + 7;
const long long N = 1e5 + 10;
const long long IM = 1e15 + 37;
const long double PI = 3.1415926535897932384;
inline void PP(long long n) { cout << n << " "; }
void PV(vector<long long> v) {
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << "\n";
}
void PVV(vector<pair<long long, long long> > v) {
  for (long long i = 0; i < v.size(); i++)
    cout << v[i].first << " " << v[i].second << "\n";
}
void PA(long long v[], long long n, long long x = 0) {
  for (long long i = x; i < n + x; i++) cout << v[i] << ' ';
  cout << "\n";
}
void IN(long long a[], long long n, long long x = 0) {
  for (long long i = x; i < n + x; i++) cin >> a[i];
}
inline void op() {}
signed main() {
  op();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  set<char> ck;
  for (char c : s) ck.insert(c);
  char c = *prev(ck.end());
  long long u = min(k, n);
  string ans = s.substr(0, u);
  if (k > n) {
    ans = s;
    while (k > n) {
      k--;
      ans += *ck.begin();
    }
    cout << (ans) << "\n";
    return 0;
  }
  bool ok = 1;
  for (long long i = k - 1; i >= 0; i--) {
    if (ans[i] != c) {
      auto it = ck.upper_bound(ans[i]);
      ans[i] = *it;
      ok = 0;
      break;
    } else {
      ans[i] = *ck.begin();
    }
  }
  if (ok) ans = s.substr(0, u);
  u = k - u;
  while (u--) ans += *ck.begin();
  cout << ans << "\n";
  return 0;
}
