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;
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
const int maxn = 1000100;
char d[100002];
int main() {
  int n, k;
  cin >> n >> k;
  map<char, int> f;
  string s;
  cin >> s;
  for (int i = 0; i < s.length(); i++) d[i] = s[i];
  sort(d, d + s.length());
  for (int i = 0; i < s.length(); i++) {
    f[d[i]] = i;
  }
  int q = s.length();
  if (k > n) {
    cout << s;
    for (int i = n; i < k; i++) {
      cout << d[0];
    }
  } else {
    s = s.substr(0, k);
    bool help = true;
    int j = s.length() - 1;
    while (true) {
      int g = f[s[j]];
      if (help) {
        g++;
        help = false;
      }
      if (g == q) {
        s[j] = d[0];
        help = true;
      } else {
        s[j] = d[g];
      }
      if (!help) break;
      j--;
    }
    cout << s;
  }
}
