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 <bits/stdc++.h>

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 = ;

// 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, , s.size() - ) {

    if (vis[s[i]] == ) {

      vec.pb(s[i]);

    }

  }

  sort(vec.begin(), vec.end());

  s1 = s;

  // cout<<len<<" "<<s1.size()<<endl;

  if (len > s1.size()) {

    cout<<s1;

    For(i, , len - s1.size())cout<<vec.front();

    return ;

  }

  s1 = s.substr(, len);

  FFor(i, len - , ) {

    char c = s1[i];

    if (c == vec.back())

      continue;

    FFor(j, vec.size() - , ) {

      if (c == vec[j]) {

        s1[i] = vec[j + ];

        For(k, i + , len - ) { s1[k] = vec.front(); }

        cout << s1;

        return ;

      }

    }

  }

  return ;

}

巴特西

C. Phone Numbers

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