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;
set<int> s;
set<int>::iterator it, it2;
map<int, int> m, n;
char z[100001];
void call(int b) {
  int k, l, g;
  if (z[b] != *it) {
    k = n[z[b]];
    z[b] = m[k + 1];
  } else {
    z[b] = *it2;
    call(b - 1);
  }
}
int main() {
  int a, b, i;
  char z2[2];
  scanf("%d %d", &a, &b);
  gets(z2);
  gets(z);
  for (i = 0; z[i] != '\0'; i++) {
    s.insert(z[i]);
  }
  for (it = s.begin(), i = 0; it != s.end(); it++, i++) {
    m[i] = *it;
    n[*it] = i;
  }
  if (b > a) {
    it = s.begin();
    for (i = 0; z[i] != '\0'; i++) printf("%c", z[i]);
    for (; i < b; i++) printf("%c", *it);
  } else {
    it = s.end();
    it--;
    it2 = s.begin();
    b--;
    call(b);
    for (i = 0; i <= b; i++) printf("%c", z[i]);
  }
}
