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 }