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 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
import sys
first=input().split()
x=int(first[0])
a=int(first[1])
b=int(first[2])
txt=input() #ya esta indexado
results=[]

def dividirParejo(txt, z, until, cant):
    print(until+cant)
    p=0
    x=0
    i=0
    while i<until:
        results.append(txt[x:p+z])
        p=p+z
        x=p
        i=i+1
    for p in results: print(p)

def aver(txt, A, B):
    if A>B:
        a=B
        b=A
    else:
        a=A
        b=B
    cant=0
    state=4 # 2 no hay solucion, 3 hay solucion, 4 sigue buscando
    inicio=0
    while  state==4:
        if a+b == (len(txt)-inicio):
            cant=cant+2
            results.append(txt[inicio:inicio+a])
            results.append(txt[inicio+a:inicio+a+b])
            state=3
        elif(len(txt)-inicio<0): #lo que me resta del txt -a <0
            state=2
            print(-1)
        elif (len(txt)-inicio)%a==0:
            state=5
            z=(len(txt)-inicio)//a
            dividirParejo(txt[inicio: len(txt)],a,z, cant)
        elif (len(txt)-inicio)%b==0:
            state=5
            z=(len(txt)-inicio)//b
            dividirParejo(txt[inicio: len(txt)],b,z, cant)
        else:
            cant=cant+1
            results.append(txt[inicio:inicio+a])
            inicio = inicio+a

        if state==3:
            print(cant)
            for p in results: print(p)

if x%a==0:
    z=x//a
    dividirParejo(txt,a, z,0)
elif x%b==0:
    z=x//b
    dividirParejo(txt,b, z,0)
else:
    aver(txt, a, b)
