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;
using ll = long long;
using ld = long double;
constexpr ll lmax = 1e18;
constexpr ll imax = 1e9;
constexpr ll imin = -1e9;
constexpr ll mod = 1e9 + 7;
constexpr ld eps = 1e-9;
int n, k;
string s, p;
void solve() {
  cin >> (n) >> (k) >> (s);
  set<char> st;
  st.insert('{');
  for (ll i = 0; i < n; i += 1) st.insert(s[i]);
  if (k > n) {
    int r = k - n;
    cout << (s);
    while (r--) cout << (*st.begin());
  } else {
    for (ll i = 0; i < k; i += 1) p += s[i];
    for (ll i = k - 1; i >= 0; i -= 1) {
      char c = p[i];
      auto ch = st.upper_bound(c);
      if (*ch == '{') {
        p[i] = *st.begin();
      } else {
        p[i] = *ch;
        break;
      }
    }
    cout << (p);
  }
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  ll tc = 1;
  while (tc--) solve();
  return 0;
}
