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;
const int N = 1000010;
vector<long long> is_prime(N + 1, true);
long long to_int(string s) {
  stringstream geek(s);
  long long first = 0;
  geek >> first;
  return first;
}
string to_string(long long n) {
  ostringstream str1;
  str1 << n;
  string geek = str1.str();
  return geek;
}
bool sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first > b.first);
}
bool sortbysecdesc(const pair<int, int> &a, const pair<int, int> &b) {
  return a.second > b.second;
}
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
long long powerm(long long a, long long k) {
  if (!k) {
    return 1;
  }
  long long b = powerm(a, k / 2);
  b = b * b % (1000000000 + 7);
  if (k % 2) {
    return a * b % (1000000000 + 7);
  } else {
    return b;
  }
}
long long power(long long a, long long b) {
  if (b == 1) return a;
  if (b == 0) return 1;
  long long m1 = power(a, b / 2);
  if (b % 2) return m1 * m1 * a;
  return m1 * m1;
}
void seive_of_eras(long long n) {
  is_prime[0] = 0;
  is_prime[1] = 0;
  for (long long i = 2; i * i <= n; i++) {
    if (is_prime[i]) {
      for (long long j = i * i; j <= n; j += i) {
        is_prime[j] = false;
      }
    }
  }
}
bool isprime(long long a) {
  if (a <= 1) return false;
  if (a == 2 | a == 3) return true;
  if (a % 2 == 0 || a % 3 == 0) return false;
  for (long long i = 5; i * i <= a; i = i + 6) {
    if (a % i == 0 || a % (i + 2) == 0) return false;
  }
  return true;
}
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  char c[100005];
  scanf("%s", c);
  map<char, int> m;
  for (int i = 0; i < n; i++) {
    m[c[i]];
  }
  int cnt = 0;
  vector<char> v;
  for (auto &it : m) {
    it.second = cnt++;
    v.push_back(it.first);
  }
  int a[100005];
  for (int i = 0; i < n; i++) {
    a[i] = m[c[i]];
  }
  if (k > n) {
    for (int i = n; i < k; i++) {
      a[i] = 0;
    }
  } else {
    int Max = m.size();
    a[k - 1]++;
    for (int i = k - 1; i >= 0; i--) {
      if (a[i] >= Max) {
        a[i] = 0;
        a[i - 1]++;
      } else {
        break;
      }
    }
  }
  for (int i = 0; i < k; i++) printf("%c", v[a[i]]);
  printf("\n");
}
