CodeForces612AThe Text Splitting（枚举）

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代码如下：

<pre name="code" class="cpp">#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char a[1100];
int main()
{int n,p,q,t;while(~scanf("%d%d%d",&n,&p,&q)){scanf("%s",a);int i,j,k,t,s=0;for(i=0;i<=n;i++){if(s!=0)break;for(j=0;j<=n;j++){if(i*p+j*q>n)break;if(i*p+j*q==n){printf("%d\n",i+j);s=1;int l=0;for(k=0;k<i*p;k++){printf("%c",a[l++]);if((k+1)%p==0)printf("\n");}for(t=0;t<j*q;t++){printf("%c",a[l++]);if((t+1)%q==0)printf("\n");}break;}}}if(!s)printf("-1\n");}}

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