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 comment(linker, "/stack:640000000")
using namespace std;
const double EPS = 1e-9;
const int INF = 0x7f7f7f7f;
const double PI = acos(-1.0);
template <class T>
inline T gcd(T a, T b) {
  return (b) == 0 ? (a) : gcd((b), ((a) % (b)));
}
template <class T>
inline T lcm(T a, T b) {
  return ((a) / gcd((a), (b)) * (b));
}
template <typename T>
string NumberToString(T Number) {
  ostringstream second;
  second << Number;
  return second.str();
}
struct debugger {
  template <typename T>
  debugger& operator,(const T& v) {
    cerr << v << " ";
    return *this;
  }
} dbg;
int t, n, m;
string s;
int mark[26];
string ans;
void add(int i) { ans += (char)(i + 'a'); }
bool dfs(int pos, int flag) {
  if (pos == m) return flag;
  if (flag == 0) {
    for (int i = 0; i < (26); i++) {
      if (mark[i] == 0) continue;
      if (pos >= n) {
        dfs(pos + 1, 1);
        add(i);
        return true;
      }
      if (s[pos] - 'a' > i) continue;
      if (s[pos] - 'a' == i) {
        if (dfs(pos + 1, 0)) {
          add(i);
          return true;
        }
      } else {
        dfs(pos + 1, 1);
        add(i);
        return true;
      }
    }
  } else {
    for (int i = 0; i < (26); i++)
      if (mark[i]) {
        dfs(pos + 1, 1);
        add(i);
        return true;
      }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  while (cin >> n >> m) {
    ans = "";
    cin >> s;
    memset(mark, 0, sizeof(mark));
    for (int i = 0; i < (n); i++) mark[s[i] - 'a'] = 1;
    dfs(0, 0);
    reverse(ans.begin(), ans.end());
    cout << ans << endl;
  }
  return 0;
}
