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;
const int N = 2055;
const int M = 1000005;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
long long fact(long long n) {
  long long ret = 1;
  for (int i = 1; i <= n; i++) ret *= i;
  return ret;
}
bool is_vowel(char c) {
  if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' || c == 'y')
    return 1;
  return 0;
}
int n, k;
string s;
bool cnt[30];
int sol[100005];
bool v[100005][2];
bool dfs(int idx, bool is) {
  if (v[idx][is]) return 0;
  v[idx][is] = 1;
  if (idx == k) return is;
  int cur = s[idx] - 'a';
  if (is) cur = 0;
  for (int i = cur; i < 26; i++) {
    if (!cnt[i]) continue;
    int newIS = is | (i > cur);
    sol[idx] = i;
    if (dfs(idx + 1, newIS)) return 1;
  }
  return 0;
}
int main() {
  cin >> n >> k >> s;
  for (int i = 0; i < n; i++) cnt[s[i] - 'a'] = 1;
  if (k > n) {
    cout << s;
    char c = 'a';
    for (int i = 0; i < 26; i++) {
      if (cnt[i]) {
        for (int j = 0; j < k - n; j++) printf("%c", c + i);
        break;
      }
    }
    return 0;
  }
  dfs(0, 0);
  for (int i = 0; i < k; i++) {
    printf("%c", 'a' + sol[i]);
  }
  return 0;
}
