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 a[600000] = {0};
long long c[600000] = {0};
long long b[600000] = {0};
int main() {
  int n, m;
  cin >> n >> m;
  string s;
  cin >> s;
  int kk = -1;
  int tt = 27;
  for (int i = 0; i < n; i++) {
    int c = s[i] - 'a';
    a[c]++;
    if (c > kk) kk = c;
    if (tt > c) tt = c;
  }
  if (m > n) {
    int l = m - n;
    while (l--) {
      char k = tt + 'a';
      s += k;
    }
    cout << s << endl;
  } else {
    string hk;
    int i;
    for (i = m - 1; i >= 0; i--) {
      int t = s[i] - 'a';
      if (t != kk) break;
    }
    if (i < 0) i = 0;
    for (int j = 0; j < i; j++) {
      hk += s[j];
    }
    int t = s[i] - 'a';
    for (int j = 0; j < 26; j++) {
      if (j > t && a[j] > 0) {
        char pp = j + 'a';
        hk += pp;
        break;
      }
    }
    for (int j = i + 1; j < m; j++) {
      char pp = tt + 'a';
      hk += pp;
    }
    cout << hk << endl;
    return 0;
  }
}
