C. Phone Numbers

http://codeforces.com/problemset/problem/940/C

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 aband 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

Copy

3 3
abc

output

aca

input

Copy

3 2
abc

output

ac

input

Copy

3 3
ayy

output

yaa

input

Copy

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 sis 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.

 

 思维题

1.s.size()>len 直接往后加最小的字符

2.其他情况，从后往前找（不是最大字符）的第一个字符，将它改为字典序下一个字符，然后将它后面的字符全改为最小字符

#include 
using namespace std;
typedef long long ll;
#define inf 2147483647
const ll INF = 0x3f3f3f3f3f3f3f3fll;
#define ri register int
template <class T> inline T min(T a, T b, T c) { return min(min(a, b), c); }
template <class T> inline T max(T a, T b, T c) { return max(max(a, b), c); }
template <class T> inline T min(T a, T b, T c, T d) {
  return min(min(a, b), min(c, d));
}
template <class T> inline T max(T a, T b, T c, T d) {
  return max(max(a, b), max(c, d));
}
#define scanf1(x) scanf("%d", &x)
#define scanf2(x, y) scanf("%d%d", &x, &y)
#define scanf3(x, y, z) scanf("%d%d%d", &x, &y, &z)
#define scanf4(x, y, z, X) scanf("%d%d%d%d", &x, &y, &z, &X)
#define pi acos(-1)
#define me(x, y) memset(x, y, sizeof(x));
#define For(i, a, b) for (int i = a; i <= b; i++)
#define FFor(i, a, b) for (int i = a; i >= b; i--)
#define bug printf("***********\n");
#define mp make_pair
#define pb push_back
const int maxn = 10005;
// name*******************************
string s, s1;
int len;
vector<char> vec;
bool vis[maxn];
int n;
// function******************************

//***************************************
int main() {
  //    ios::sync_with_stdio(0);
  //    cin.tie(0);
  // freopen("test.txt", "r", stdin);
  //  freopen("outout.txt","w",stdout);
  cin >> n >> len;
  cin>>s;
  For(i, 0, s.size() - 1) {
    if (vis[s[i]] == 0) {
      vec.pb(s[i]);
    }
  }
  sort(vec.begin(), vec.end());
  s1 = s;
  // cout<
  if (len > s1.size()) {
    cout<<s1;
    For(i, 1, len - s1.size())cout<<vec.front();
    return 0;
  }
  s1 = s.substr(0, len);
  FFor(i, len - 1, 0) {
    char c = s1[i];
    if (c == vec.back())
      continue;
    FFor(j, vec.size() - 2, 0) {
      if (c == vec[j]) {
        s1[i] = vec[j + 1];
        For(k, i + 1, len - 1) { s1[k] = vec.front(); }
        cout << s1;
        return 0;
      }
    }
  }

  return 0;
}

 

 

转载于:https://www.cnblogs.com/planche/p/8627423.html