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>
#pragma GCC optimize(2)
using namespace std;
const int INF = 1e9 + 7;
const double eps = 1e-6;
const int MAXN = 1e6 + 7;
int n, k;
char s[MAXN];
bool vis[30];
int main() {
  int l = INT_MAX, r = INT_MIN;
  cin >> n >> k;
  cin >> s;
  for (int i = 0; s[i]; i++) {
    vis[s[i] - 'a'] = 1;
    l = min(l, s[i] - 'a');
    r = max(r, s[i] - 'a');
  }
  if (n < k) {
    cout << s;
    for (int i = n + 1; i <= k; i++) cout << char(l + 'a');
    cout << endl;
  } else {
    int i, j;
    for (i = k - 1; ~i; i--)
      if (r + 'a' > s[i]) {
        for (int t = 0; t < i; t++) cout << s[t];
        for (j = s[i] - 'a' + 1; j <= r; j++)
          if (vis[j]) {
            cout << char(j + 'a');
            break;
          }
        for (j = i + 1; j < k; j++) cout << char(l + 'a');
        cout << endl;
        break;
      }
  }
  return 0;
}
