Problem: Find out if it is possible to partition the first n positive integers into two non-empty disjoint sets S_1 and S_2 such that:

gcd(sum(S_1), sum(S_2)) > 1 

Here sum(S) denotes the sum of all elements present in set S and gcd means the[greatest common divisor](https://en.wikipedia.org/wiki/Greatest_common_divisor).

Every integer number from 1 to n should be present in exactly one of S_1 or S_2.

Input

The only line of the input contains a single integer n (1 ≤ n ≤ 45 000)

Output

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

Otherwise, print "Yes" (quotes for clarity), followed by two lines, describing S_1 and S_2 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 2 and 4 respectively. The gcd(2, 4) = 2 > 1, hence that is one of the possible answers.

Solution (in JAVA):

Solution:
//package com.jetbrains;

import java.io.*;
import java.util.*;
import java.lang.*;
public class Main {
	public static void main(String[] args) throws FileNotFoundException {
		try
		{
			InputReader in = new InputReader(System.in);
			PrintWriter out = new PrintWriter(System.out);
			int n=in.nextInt();
			if(n<=2)
				out.print("No");
			else {
				int e = n / 2;
				int o = n - e;
				out.println("Yes");
				out.print(o + " " );
				for (int i = 1; i <= n; i += 2)
					out.print(i + " ");
				out.println();
				out.print(e+" ");
				for (int i = 2; i <= n; i += 2)
					out.print(i + " ");
			}
			out.close();
		}
		catch(Exception e){

		}
	}
	public static long power(long x, long y, long p)
	{
		// Initialize result
		long res = (long)1;

		// Update x if it is more
		// than or equal to p
		x = x % p;

		while (y > 0)
		{
			// If y is odd, multiply x
			// with result
			if((y & 1)==1)
				res = (res * x) % p;

			// y must be even now
			// y = y / 2
			y = y >> 1;
			x = (x * x) % p;
		}
		return res;
	}
	public static void merge(int arr[], int l, int m, int r)
	{
		int n1 = m - l + 1;
		int n2 = r - m;
		int L[] = new int [n1];
		int R[] = new int [n2];
		for (int i=0; i<n1; ++i)
			L[i] = arr[l + i];
		for (int j=0; j<n2; ++j)
			R[j] = arr[m + 1+ j];
		int i = 0, j = 0;
		int k = l;
		while (i < n1 && j < n2)
		{
			if (L[i] <= R[j])
			{
				arr[k] = L[i];
				i++;
			}
			else
			{
				arr[k] = R[j];
				j++;
			}
			k++;
		}
		while (i < n1)
		{
			arr[k] = L[i];
			i++;
			k++;
		}
		while (j < n2)
		{
			arr[k] = R[j];
			j++;
			k++;
		}
	}
	public static void sort(int arr[], int l, int r)
	{
		if (l < r)
		{
			int m = (l+r)/2;
			sort(arr, l, m);
			sort(arr , m+1, r);
			merge(arr, l, m, r);
		}
	}
	static class InputReader {

		private final InputStream stream;
		private final byte[] buf = new byte[8192];
		private int curChar, snumChars;

		public InputReader(InputStream stream) {
			this.stream = stream;
		}

		public int read() {
			if (snumChars == -1)
				throw new InputMismatchException();
			if (curChar >= snumChars) {
				curChar = 0;
				try {
					snumChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (snumChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		public int nextInt() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public long nextLong() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public int[] nextIntArray(int n) {
			int a[] = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = nextInt();
			}
			return a;
		}

		public String readString() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public String nextLine() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndOfLine(c));
			return res.toString();
		}

		public boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		private boolean isEndOfLine(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}

	}
}