CodeForces 544A

You are given a string q. A sequence of k strings s₁, s₂, ..., s_k is called beautiful, if the concatenation of these strings is string q (formally, s₁ + s₂ + ... + s_k = q) and the first characters of these strings are distinct.

Find any beautiful sequence of strings or determine that the beautiful sequence doesn't exist.

Input

The first line contains a positive integer k (1 ≤ k ≤ 26) — the number of strings that should be in a beautiful sequence.

The second line contains string q, consisting of lowercase Latin letters. The length of the string is within range from 1 to 100, inclusive.

Output

If such sequence doesn't exist, then print in a single line "NO" (without the quotes). Otherwise, print in the first line "YES" (without the quotes) and in the next k lines print the beautiful sequence of strings s₁, s₂, ..., s_k.

If there are multiple possible answers, print any of them.

Sample test(s)

Input

1
abca

Output

YES
abca

Input

2
aaacas

Output

YES
aaa
cas

Input

4
abc

Output

NO

Note

In the second sample there are two possible answers: {"aaaca", "s"} and {"aaa", "cas"}.

做简单题目，慢慢来，想快点，写好点

 

 1 #include <cstring>

 2 #include <cstdio>

 3 #include <iostream>

 4 #include <set>

 5 using namespace std;

 6 

 7 int main()

 8 {

 9     //freopen("in.txt", "r", stdin);

10     int nstr;

11 

12     while(scanf("%d", &nstr) != EOF)

13     {

14         getchar();

15         char str[105];

16         bool vis[105];

17         gets(str);

18         int len = strlen(str);

19         memset(vis, false, sizeof(vis));

20 

21         set<char> s;

22 

23         int p = 0;

24         s.insert(str[0]);

25         ++p;

26         vis[0] = true;

27         if(p != nstr)

28         {

29             for(int i = 1; i != len; ++i)

30             {

31                 if(s.find(str[i]) == s.end() && p < nstr)

32                 {

33                     s.insert(str[i]);

34                     ++p;

35                     vis[i] = true;

36                 }

37             }

38         }

39 

40         if(p < nstr)

41         {

42             cout << "NO" << endl;

43         }

44         else

45         {

46             cout << "YES" << endl;

47             for(int i = 0; i != len; ++i)

48             {

49                 if(vis[i] == true)

50                 {

51                     putchar(str[i]);

52                     int j = i+1;

53                     while(vis[j] == false && j < len)

54                     {

55                         putchar(str[j]);

56                         ++j;

57                     }

58                     puts("");

59                 }

60             }

61         }

62     }

63     return 0;

64 }

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