Levko loves permutations very much. A permutation of length n is a sequence of distinct positive integers, each is at most n.

Let’s assume that value gcd(a, b) shows the greatest common divisor of numbers a and b. Levko assumes that element pi of permutation p1, p2, ... , pn is good if gcd(i, pi) > 1. Levko considers a permutation beautiful, if it has exactly k good elements. Unfortunately, he doesn’t know any beautiful permutation. Your task is to help him to find at least one of them.

Input

The single line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ n).

Output

In a single line print either any beautiful permutation or -1, if such permutation doesn’t exist.

If there are multiple suitable permutations, you are allowed to print any of them.

Examples

Input

4 2


Output

2 4 3 1

Input

1 1


Output

-1

Note

In the first sample elements 4 and 3 are good because gcd(2, 4) = 2 > 1 and gcd(3, 3) = 3 > 1. Elements 2 and 1 are not good because gcd(1, 2) = 1 and gcd(4, 1) = 1. As there are exactly 2 good elements, the permutation is beautiful.

The second sample has no beautiful permutations.
function trim(s) {
	return s.replace(/^\s+|\s+$/gm, '');
}

function tokenize(s) {
	return trim(s).split(/\s+/);
}

function tokenizeIntegers(s) {
	var tokens = tokenize(s);
	for (var i = 0; i < tokens.length; i += 1) {
		tokens[i] = parseInt(tokens[i]);
	};
	return tokens;
}

function main() {
	var integers = tokenizeIntegers(readline());
	var n = integers[0], good = integers[1];

	if (good == n) {
		print(-1);
		return;
	}

	var perm = [];
	for (var i = 1; i <= n; ++i) {
		perm.push(i);
	}

	var bad = n-good;
	for (var pos = 1; pos < bad; ++pos) {
		perm[pos-1] = pos+1;
		perm[pos] = 1;
	}
	print(perm.join(" "));
}

main();
