题干：

Find out if it is possible to partition the first nn positive integers into two non-empty disjoint sets S1S1 and S2S2 such that:

gcd(sum(S1),sum(S2))>1gcd(sum(S1),sum(S2))>1

Here sum(S)sum(S) denotes the sum of all elements present in set SS and gcdgcd means thegreatest common divisor.

Every integer number from 11 to nn should be present in exactly one of S1S1 or S2S2.

Input

The only line of the input contains a single integer nn (1≤n≤450001≤n≤45000)

Output

If such partition doesn't exist, print "No" (quotes for clarity).

Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing S1S1and S2S2 respectively.

Each set description starts with the set size, followed by the elements of the set in any order. Each set must be non-empty.

If there are multiple possible partitions — print any of them.

Examples

Input

1

Output

No

Input

3

Output

Yes
1 2
2 1 3

Note

In the first example, there is no way to partition a single number into two non-empty sets, hence the answer is "No".

In the second example, the sums of the sets are 22 and 44 respectively. The gcd(2,4)=2>1gcd(2,4)=2>1, hence that is one of the possible answers.

题目大意：

给一个数n，将1到n之间的整数，分成两个集合，每组的和分别为sum1,sum2，要求这个集合的gcd(sum1,sum2)>1，输出每组的大小和每一个元素。

解题报告：

   乱搞一下就可以了。，

AC代码：

#include<bits/stdc++.h>

using namespace std;
const int MAX = 2e5 + 5;
int main()
{
	int n;
	cin>>n;
	if(n <= 2) {
		puts("No");return 0 ;
	}
	puts("Yes");
	int s1 = (n+1)/2;
	printf("1 %d\n",s1);
	printf("%d",n-1);
	for(int i = 1; i<=n; i++) {
		if(i == s1) continue;
		printf(" %d",i);
	}
	return 0 ;
}

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