codeforces_612A. The Text Splitting

A. The Text Splitting

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

5 2 3Hello

output

2Hello

input

10 9 5Codeforces

output

2Codeforces

input

6 4 5Privet

output

-1

input

8 1 1abacabac

output

8abacabac

一道水题wa了两次，因为没有考虑到p和q成不同倍数与n相等的情况。。好吧状态很差代码很丑。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>using namespace std;char aa[110];int a,b;int solve(int p,int q,int x){    int flag=0;    int qq=q,pp=p;    for(a=0;p+q<=100;a++)    {        p=pp;        p*=a;        for(b=0;p+q<=100;b++)        {            q=qq;            q*=b;            if(p+q==x)            {                flag=1;break;            }        }        if(flag)break;        q=0;    }    return flag;}int main(){    int n,p,q,i,j;    scanf("%d%d%d",&n,&p,&q);    scanf("%s",&aa);    int len=strlen(aa);    if(p+q==len)    {        printf("%d\n",2);        for(i=0;i<p;i++)            printf("%c",aa[i]);        printf("\n");        for(;i<len;i++)            printf("%c",aa[i]);        printf("\n");    }    else if(solve(p,q,n))    {        int l=0;        printf("%d\n",a+b);        for(i=0;i<a;i++)        {            for(j=0;j<p;j++)            {                printf("%c",aa[l]);                l++;            }            printf("\n");        }        for(i=0;i<b;i++)        {            for(j=0;j<q;j++)            {                printf("%c",aa[l]);                l++;            }            printf("\n");        }    }    else    {        printf("-1\n");    }    return 0;}