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 ull = unsigned long long;
const int MOD = 1e9 + 7;
const int N = 2e5 + 5;
const ll INF = 9e18;
const int M = 501;
string mn(string a, string b) {
  if ((int)a.size() > (int)b.size())
    return b;
  else if ((int)a.size() < (int)b.size())
    return a;
  else {
    for (int i = 0; i < (int)a.size(); ++i) {
      if (a[i] > b[i])
        return b;
      else if (a[i] < b[i])
        return a;
    }
    return a;
  }
}
bool have[N];
void solve() {
  int n, m;
  cin >> n >> m;
  string s;
  cin >> s;
  string ans = "";
  string md = "";
  int Mn = 27;
  char x;
  for (int i = 0; i < n; ++i) {
    if (Mn > s[i] - 'a') {
      Mn = s[i] - 'a';
      x = s[i];
    }
    have[s[i] - 'a'] = 1;
  }
  for (int i = 0; i < m; ++i) md += x;
  for (int i = 0; i < m; ++i) {
    int pos = -1;
    string cur = md;
    int y = s[i] - 'a';
    ++y;
    if (min(cur, s) != cur) {
      if ((int)ans.size() == 0)
        ans = cur;
      else
        ans = min(ans, cur);
      if (i < (int)s.size()) md[i] = s[i];
      continue;
    }
    if (y >= 0) {
      for (int j = y; j < 26; ++j) {
        if (have[j]) {
          pos = j;
          break;
        }
      }
    }
    if (pos == -1) {
      if (i < (int)s.size())
        md[i] = s[i];
      else
        continue;
      cur = md;
      continue;
    } else
      cur[i] = char(pos + 'a');
    md[i] = s[i];
    if (min(cur, s) == cur) continue;
    ans = cur;
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) solve();
}
