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 3Hello
Output
2Hello
Input
10 9 5Codeforces
Output
2Codeforces
Input
6 4 5Privet
Output
-1
Input
8 1 1abacabac
Output
8abacabac
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 if(n % p == 0 || n % q == 0)
11 {
12 l1=(n%p == 0)?p:q;
13 l2=n/l1;
14 k=-1;
15 printf("%d\n",l2);
16 while(l2--)
17 {
18 l=l1;
19 while(l--)
20 {
21 printf("%c",s[++k]);
22 }
23 printf("\n");
24 }
25 }
26 else
27 {
28 for(int i = 1; i < n ; i++)
29 {
30 for( int j = 1; j < n ; j++)
31 {
32 if(i*p+j*q == n)
33 {
34 printf("%d\n",i+j);
35 k=-1;
36 while(i--)
37 {
38 l=p;
39 while(l--)
40 {
41 printf("%c",s[++k]);
42 }
43 printf("\n");
44 }
45
46 while(j--)
47 {
48 l=q;
49 while(l--)
50 {
51 printf("%c",s[++k]);
52 }
53 printf("\n");
54 }
55 g=0;
56 break;
57 }
58 }
59 if(g == 0) break;
60 }
61 if(g != 0) printf("-1\n");
62 }
63
64
65 }
66
67 }