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;
long long int ll_max(long long int a, long long int b, long long int c) {
  return max(a, max(b, c));
}
int int_max(int a, int b, int c) { return max(a, max(b, c)); }
long long int ll_min(long long int a, long long int b, long long int c) {
  return min(a, min(b, c));
}
int int_min(int a, int b, int c) { return min(a, min(b, c)); }
long long int max(int a, long long int b) { return max((long long int)a, b); }
long long int min(int a, long long int b) { return min((long long int)a, b); }
long long int min(long long int a, int b) { return min(a, (long long int)b); }
long long int max(long long int a, int b) { return max(a, (long long int)b); }
long long int dx[] = {0, 0, 1, -1};
long long int dy[] = {1, -1, 0, 0};
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << ' ' << *it << " = " << a;
  err(++it, args...);
}
template <typename T, typename U>
inline void min_self(T& x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void max_self(T& x, U y) {
  if (x < y) x = y;
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> pair) {
  return out << "(" << pair.first << ", " << pair.second << ")";
}
template <class T>
ostream& operator<<(ostream& out, vector<T> vec) {
  out << "(";
  for (auto& v : vec) out << v << ", ";
  return out << ")";
}
template <class T>
ostream& operator<<(ostream& out, set<T> vec) {
  out << "(";
  for (auto& v : vec) out << v << ", ";
  return out << ")";
}
template <class L, class R>
ostream& operator<<(ostream& out, map<L, R> vec) {
  out << "(";
  for (auto& v : vec) out << "[" << v.first << ", " << v.second << "]";
  return out << ")";
}
template <class A, class B>
istream& operator>>(istream& in, pair<A, B>& a) {
  return in >> a.first >> a.second;
}
template <class A>
istream& operator>>(istream& in, vector<A>& a) {
  for (A& i : a) in >> i;
  return in;
}
void solve() {
  long long int n, k;
  cin >> n >> k;
  string s, t;
  cin >> s;
  set<char> c;
  for (long long int i = 0; i < n; i++) c.insert(s[i]);
  if (k > n) {
    cout << s;
    for (long long int i = 0; i < k - n; i++) cout << *c.begin();
    cout << '\n';
    return;
  }
  t = s.substr(0, k);
  auto it = c.end();
  it--;
  for (long long int i = k - 1; i >= 0; i--) {
    if (t[i] != *it) {
      it = c.lower_bound(t[i]);
      it++;
      t[i] = *it;
      for (long long int j = i + 1; j < k; j++) {
        t[j] = *c.begin();
      }
      break;
    }
  }
  cout << t << '\n';
  return;
}
int main() {
  (ios::sync_with_stdio(false), cin.tie(0));
  long long int t = 1;
  while (t--) {
    solve();
  }
}
