The Text Splitting

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

#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>char s[101]; int main(){    int n;    int q,p;    scanf("%d%d%d",&n,&q,&p);   scanf("%s",s);      if(n%q==0){printf("%d\n",n/q);for(int j=0;j<(n/q);j++){for(int i=j*q;i<(j+1)*q;i++)       {   printf("%c",s[i]);       }       printf("\n");    }     return 0;}   if(n%p==0)    {  printf("%d\n",n/p);  for(int j=0;j<(n/p);j++)  {for(int i=j*p;i<(j+1)*p;i++)       {   printf("%c",s[i]);       }       printf("\n");     }     return 0;    } for( int i=1;i<=100;i++) { for(int j=1;j<=100;j++) { if(n==i*q+p*j) { printf("%d\n",i+j); for(int x=0;x<i;x++) { for(int y=x*q;y<(x+1)*q;y++) { printf("%c",s[y]); } printf("\n"); } for(int x=0;x<j;x++) { for(int y=x*p+q*i;y<(x+1)*p+q*i;y++)// 例如13==5+5+3；y的id要从q*i开始。 { printf("%c",s[y]); } printf("\n"); } return 0; } } }    printf("-1\n");return 0;}