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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  t = 1;
  while (t--) {
    long long n, k, i, j;
    cin >> n >> k;
    char ch = 'z';
    vector<long long> v(26, -1);
    string s, a = "";
    cin >> s;
    ch = 'z';
    for (i = 0; i < n; i++) {
      if (ch > s[i]) {
        ch = s[i];
      }
    }
    if (n < k) {
      cout << s;
      long long y = k - n;
      while (y--) {
        cout << ch;
      }
    } else {
      for (i = 0; i < n; i++) {
        v[s[i] - 'a'] = i;
      }
      long long mx = -1, f = 0, id = -1;
      set<char> st;
      a = s;
      if (f) {
        for (i = 0; i < k; i++) cout << a[i];
      } else {
        set<char> ss;
        char cc = 'z';
        for (i = k - 1; i >= 0; i--) {
          for (j = 0; j < 26; j++) {
            char d = 'a' + j;
            if (s[i] < d && v[d - 'a'] != -1) {
              mx = i;
              cc = d;
              break;
            }
          }
          if (mx != -1) {
            break;
          }
        }
        a[mx] = cc;
        for (i = mx + 1; i < k; i++) a[i] = ch;
        for (i = 0; i < k; i++) cout << a[i];
      }
    }
  }
  return 0;
}
