The Text Splitting (将字符串分成若干份,每份长度为p或q)

Description

You are given the string s of length n and the numbers p, q. Split the string s to pieces of length p and q.

For example, the string "Hello" for p = 2, q = 3 can be split to the two strings "Hel" and "lo" or to the two strings "He" and "llo".

Note it is allowed to split the string s to the strings only of length p or to the strings only of length q (see the second sample test).

Input

The first line contains three positive integers n, p, q (1 ≤ p, q ≤ n ≤ 100).

The second line contains the string s consists of lowercase and uppercase latin letters and digits.

Output

If it's impossible to split the string s to the strings of length p and q print the only number "-1".

Otherwise in the first line print integer k — the number of strings in partition of s.

Each of the next k lines should contain the strings in partition. Each string should be of the length p or q. The string should be in order of their appearing in string s — from left to right.

If there are several solutions print any of them.

Sample Input

Input

5 2 3
Hello

Output

2
He
llo

Input

10 9 5
Codeforces

Output

2
Codef
orces

Input

6 4 5
Privet

Output

-1

Input

8 1 1
abacabac

Output

8
a
b
a
c
a
b
a
c
 1 #include<cstdio>
 2 char s[1000000];
 3 int main()
 4 {
 5     int n,p,q,k,l1,l2,l,g;
 6     while(scanf("%d %d %d",&n,&p,&q)!=EOF)
 7     {
 8         scanf("%s",&s);
 9         g=1;
10         for(int i = 0; i <= n ; i++)
11             {
12                  for( int j = 0; j <= n ; j++)
13                 {
14                     if(i*p+j*q == n)
15                     {
16                         printf("%d
",i+j);
17                         k=-1;
18                         while(i--)
19                         {
20                             l=p;
21                             while(l--)
22                             {
23                                 printf("%c",s[++k]);
24                             }
25                             printf("
");
26                         }        
27                         while(j--)
28                         {
29                             l=q;
30                             while(l--)
31                             {
32                                 printf("%c",s[++k]);
33                             }
34                             printf("
");
35                         }
36                         g=0;
37                         break;
38                     }
39                 }
40                 if(g == 0) break;
41             }
42             if(g != 0) printf("-1
");
43             
44     }
45     
46 }
——将来的你会感谢现在努力的自己。
原文地址:https://www.cnblogs.com/yexiaozi/p/5692699.html