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
# Problem: E. Anfisa the Monkey
# Contest: Codeforces - School Team Contest #2 (Winter Computer School 2010/11)
# URL: https://codeforces.com/problemset/problem/44/E
# Memory Limit: 256 MB
# Time Limit: 2000 ms
# 
# KAPOOR'S

from sys import stdin, stdout 

def INI():
	return int(stdin.readline())
	
def INL():
	return [int(_) for _ in stdin.readline().split()]
	
def INS():
	return stdin.readline()

def MOD():
    return pow(10,9)+7
	
def OPS(ans):
	stdout.write(str(ans)+"\n")
	
def OPL(ans):
	[stdout.write(str(_)+" ") for _ in ans]
	stdout.write("\n")

	
if __name__=="__main__":
	k,a,b=INL()
	S=INS()
	n=len(S)-1
	if k*a<=n<=k*b:
		i=0
		for _ in range(k):
			OPS(S[i:i+n//(k-_)])
			i+=n//(k-_)
			n-=n//(k-_)
	else:
		OPS("No solution")