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>
const long long int MOD = 1000000007ll;
const int mxsz = 200005;
using namespace std;
double dist(double x1, double y1, double x2, double y2) {
  return (double)sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
int main() {
  std::ios_base::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  string st;
  cin >> st;
  int ar[26] = {};
  int mn = INT_MAX;
  int mx = 0;
  for (int i = 0; i < n; i++) {
    ar[st[i] - 'a'] = 1;
    mn = min(mn, st[i] - 'a');
    mx = max(mx, st[i] - 'a');
  }
  if (k > n) {
    cout << st;
    k -= n;
    char ch = mn + 'a';
    while (k--) {
      cout << ch;
    }
  } else {
    int i = k - 1;
    char ch = mx + 'a';
    while (1) {
      if (st[i] == ch) {
        i--;
      } else {
        break;
      }
    }
    for (int j = 0; j < i; j++) {
      cout << st[j];
    }
    for (int j = st[i] - 'a' + 1; j < 26; j++) {
      if (ar[j] > 0) {
        char ch = j + 'a';
        cout << ch;
        break;
      }
    }
    ch = mn + 'a';
    for (int j = i + 1; j < k; j++) {
      cout << ch;
    }
  }
  return 0;
}
