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;
inline int read() {
  int x = 0, f = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = -1;
  } while (ch < '0' || ch > '9');
  do {
    x = x * 10 + ch - '0';
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  return x * f;
}
const int N = 100010;
char ch[N];
bool appear[26];
int nxt[26];
inline char getnext(char x) {
  x -= 'a';
  if (nxt[x] != -1) return nxt[x] + 'a';
  nxt[x] = (x + 1) % 26;
  while (!appear[nxt[x]]) nxt[x] = (nxt[x] + 1) % 26;
  return nxt[x] + 'a';
}
int main() {
  int n = read(), k = read();
  scanf("%s", ch + 1);
  for (int i = 0; i < 26; ++i) nxt[i] = -1;
  for (int i = 1; i <= n; ++i) {
    appear[ch[i] - 'a'] = 1;
  }
  int flag = 1;
  ch[k + 1] = 0;
  if (k > n) {
    int mn = 0;
    while (!appear[mn]) ++mn;
    for (int i = n + 1; i <= k; ++i) ch[i] = mn + 'a';
    printf("%s", ch + 1);
    return 0;
  }
  for (int i = k; i >= 1; --i) {
    if (!flag) break;
    char v = getnext(ch[i]);
    if (v > ch[i]) flag = 0;
    ch[i] = v;
  }
  printf("%s", ch + 1);
  return 0;
}
