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;
bool isprime(int64_t n) {
  for (int64_t i = 2; i * i <= n; ++i) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
int64_t factorial(int64_t n) {
  return (n == 1 || n == 0) ? 1 : n * factorial(n - 1);
}
int64_t gcd(int64_t a, int64_t b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int64_t lcm(int64_t a, int64_t b) { return (a / gcd(a, b) * b); }
int64_t max(int64_t a, int64_t b) {
  int64_t sol = a > b ? a : b;
  return sol;
}
int64_t min(int64_t a, int64_t b) {
  int64_t sol = a < b ? a : b;
  return sol;
}
int64_t power(int64_t x, int64_t y) {
  int64_t ans = 1;
  while (y > 0) {
    if (y & 1) {
      ans *= x;
    }
    y >>= 1LL;
    x *= x;
  }
  return ans;
}
int64_t inf = 100000000001;
int64_t x[100005] = {0};
inline int64_t solve() {
  int64_t n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  vector<char> v;
  for (int64_t i = 0; i < n; i++) {
    v.push_back(s[i]);
  }
  sort((v).begin(), (v).end());
  string ans = "";
  if (k > n) {
    ans = s;
    for (int64_t i = n; i < k; i++) ans += v[0];
  } else {
    bool flag = true;
    for (int64_t i = k - 1; i >= 0; i--) {
      if (flag == true) {
        int64_t id = upper_bound((v).begin(), (v).end(), s[i]) - v.begin();
        if (id == v.size()) {
          ans += v[0];
        } else {
          flag = false;
          ans += v[id];
        }
      } else
        ans += s[i];
    }
    reverse((ans).begin(), (ans).end());
  }
  cout << ans;
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  solve();
}
