You are given a string q. A sequence of k strings s1, s2, ..., sk is called beautiful, if the concatenation of these strings is string q (formally, s1 + s2 + ... + sk = 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, ..., sk.
If there are multiple possible answers, print any of them.
Sample test(s)
Input
1abca
Output
YESabca
Input
2aaacas
Output
YESaaacas
Input
4abc
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 }