Permutation *p* is an ordered set of integers *p*<sub class="lower-index">1</sub>,  *p*<sub class="lower-index">2</sub>,  ...,  *p*<sub class="lower-index">*n*</sub>, consisting of *n* distinct positive integers, each of them doesn't exceed *n*. We'll denote the *i*-th element of permutation *p* as *p*<sub class="lower-index">*i*</sub>. We'll call number *n* the size or the length of permutation *p*<sub class="lower-index">1</sub>,  *p*<sub class="lower-index">2</sub>,  ...,  *p*<sub class="lower-index">*n*</sub>.

The decreasing coefficient of permutation *p*<sub class="lower-index">1</sub>, *p*<sub class="lower-index">2</sub>, ..., *p*<sub class="lower-index">*n*</sub> is the number of such *i* (1 ≤ *i* &lt; *n*), that *p*<sub class="lower-index">*i*</sub> &gt; *p*<sub class="lower-index">*i* + 1</sub>.

You have numbers *n* and *k*. Your task is to print the permutation of length *n* with decreasing coefficient *k*.
The single line contains two space-separated integers: *n*, *k* (1 ≤ *n* ≤ 10<sup class="upper-index">5</sup>, 0 ≤ *k* &lt; *n*) — the permutation length and the decreasing coefficient.
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# Richard Mello
# B - Slightly Decreasing Permutations 
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# Função principal
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def main():
	# Recebe
	n, k = [int(i) for i in input().split()]

	# Mostra a resposta
	vez = 1
	ate = n
	cont = 0
	
	while vez <= ate:
		# Se é pra inverter:
		if cont < k:
			# Mostra o ultimo
			print(ate, end = ' ')
			# Diminui o limite
			ate -= 1
			# E acrescenta o contador
			cont += 1
		# Se não é
		else:
			# Mostra o próximo
			print(vez, end = ' ')
			# Sobe
			vez += 1
	
	print()

main()

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