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 != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
  return (a / gcd<T>(a, b) * b);
}
template <class T>
inline bool read(T &x) {
  int c = getchar();
  int sgn = 1;
  while (~c && c < '0' || c > '9') {
    if (c == '-') sgn = -1;
    c = getchar();
  }
  for (x = 0; ~c && '0' <= c && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  x *= sgn;
  return ~c;
}
const int INF = 2000000009;
const int MX = 100005;
const double EPS = 1e-9;
const int MOD = 1000000007;
int main() {
  int n = ({
    int a;
    read(a);
    a;
  }),
      k = ({
        int a;
        read(a);
        a;
      });
  char s[MX];
  scanf("%s", s);
  char mn = 'z', mx = 'a';
  for (int i = 0; i < n; i++) {
    mn = min(mn, s[i]);
    mx = max(mx, s[i]);
  }
  if (n < k) {
    printf("%s", s);
    while (n < k) {
      printf("%c", mn);
      ;
      n++;
    }
    printf("\n");
    return 0;
  }
  int ind = -1;
  for (int i = 0; i < min(k, n); i++)
    if (s[i] != mx) ind = i;
  char temp = 'z';
  for (int i = 0; i < n; i++)
    if (s[i] > s[ind]) temp = min(temp, s[i]);
  s[ind++] = temp;
  while (ind < k) s[ind++] = mn;
  s[ind] = '\0';
  printf("%s\n", s);
  return 0;
}
