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;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pci = pair<char, int>;
using vi = vector<int>;
using vll = vector<ll>;
const ll infll = 1e18 + 3;
const int maxn = 1e6 + 77;
const ll basell = 1e18 + 3;
const int mod = 1e9 + 7;
const ld eps = 1e-7;
const int inf = 1009000999;
const int baseint = 1000200013;
const ld PI = acos(-1.0);
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
int inline solve() {
  ios::sync_with_stdio(NULL), cin.tie(NULL), cout.tie(NULL);
  ;
  int n, k;
  cin >> n >> k;
  string str;
  cin >> str;
  char a = 'z', b = 'a';
  set<char> h;
  for (int i = 0; i < n; ++i) {
    h.insert(str[i]);
    a = min(str[i], a);
    b = max(str[i], b);
  }
  if (n < k) {
    for (int j = 0; j < n; ++j) {
      cout << str[j];
    }
    for (int i = n; i < k; ++i) {
      cout << a;
    }
    return 0;
  }
  int o = 0;
  for (int i = n - 1; i >= 0; --i) {
    if (str[i] < b && i < k) {
      o = 1;
      for (int j = 0; j < i; ++j) {
        cout << str[j];
      }
      for (auto it = h.begin(); it != h.end(); ++it) {
        if (*it > str[i]) {
          cout << *it;
          break;
        }
      }
      for (int j = i + 1; j < k; ++j) {
        cout << a;
      }
      break;
    }
  }
  if (o == 0) {
    for (int j = 0; j < n; ++j) {
      cout << str[j];
    }
  }
}
int32_t main() { solve(); }
