Anfisa the monkey learns to type. She is yet unfamiliar with the "space" key and can only type in lower-case Latin letters. Having typed for a fairly long line, Anfisa understood that it would be great to divide what she has written into k lines not shorter than a and not longer than b, for the text to resemble human speech more. Help Anfisa.

Input

The first line contains three integers k, a and b (1 ≤ k ≤ 200, 1 ≤ a ≤ b ≤ 200). The second line contains a sequence of lowercase Latin letters — the text typed by Anfisa. It is guaranteed that the given line is not empty and its length does not exceed 200 symbols.

Output

Print k lines, each of which contains no less than a and no more than b symbols — Anfisa's text divided into lines. It is not allowed to perform any changes in the text, such as: deleting or adding symbols, changing their order, etc. If the solution is not unique, print any of them. If there is no solution, print "No solution" (without quotes). 

Examples

Input

3 2 5
abrakadabra


Output

ab
rakad
abra


Input

4 1 2
abrakadabra


Output

No solution






k,a,b= map(int,input().split())


import math
t = input()


s=len(t)

#print(s)

if a>s:
    print('No solution')
elif a==s:
    print(t)
elif a>math.ceil(s/k):
    print('No solution')
elif math.ceil(s/k)>b:
    print('No solution')
elif k>s:
    print('No solution')
else:



    if math.ceil(s/k) ==k:
        for j in range(0,s,k):
            print(t[j:j+k])
    else:
        if a==b:
            print('No solution')
        else:
            l=[s//k]*k

            x = s-((s//k)*k)

            for j in range(x):
                l[j]+=1

            bg=[]
            hg=0
            if len(l)==1:
                pass
            else:
                for j in range(1,len(l)):
                    l[j]=l[j-1]+l[j]

                for g in range(len(l)):
                    if g==0:
                        if a<=len(t[:l[g]])<=b:
                            bg.append(t[:l[g]])
                        else:
                            print('No solution')
                            hg+=1
                            break
                    else:
                        if a<=len(t[l[g-1]:l[g]])<=b:
                            bg.append(t[l[g-1]:l[g]])
                        else:
                            print('No solution')
                            hg+=1
                            break
                        
                if hg==0:
                    for y in bg:
                        print(y)
                    

        
        
