Code Forces 612A-----思维题

The Text Splitting

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

长度为n的字符串，分为x个长度为p和y个长度为q的子串，如果有多种分法，输出任意一种。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char s[10010], ss[1010];int main(){int n, p, q, a, t, k;while(~scanf("%d%d%d", &n, &p, &q)){scanf(" %s", s);k = 0;if(n % p == 0){//也可以换成qprintf("%d\n", n / p);for(int i = 0; i < n; i++){printf("%c", s[i]);if((i+1) % p == 0){printf("\n");}}continue;}else{a = 0;while(a < n){t = n;t -= a;if(t % q == 0){//n减去x个p之后可以被q整除，则可以均分printf("%d\n", t / q + a / p);for(int i = 0; i < n; i++){printf("%c", s[i]);if((i+1) % p == 0 && i < a - 1){printf("\n");}if((i+1-a) % q == 0 && i >= a - 1){printf("\n");}}k++;break;}a += p;}if(k){continue;}}printf("-1\n");//不能均分输出-1}return 0;}