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:
import java.io.*;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Non_Coprime {
    public static void main(String[] args) {

        FastReader in = new FastReader();
        OutputStream outputStream = System.out;
        PrintWriter out = new PrintWriter(outputStream);
        int i,j,a,b,c,n,k= 0,l,m;
        boolean is = false,is1 =false;
        long sum = 0;
        int[] brr = new int[1000];
        int[] arr = new int[10000];

//
//        for (i=2; i<= Math.sqrt(10313); i++)
//            if (10313 % i == 0) {
//            out.println("NOT");
//                out.flush();
//                return;
//            }

        n = in.nextInt();

        for (i=3; i<45000; i++){

            a = (int)Math.sqrt(i);
            is = false;

            for (j=2; j<=a; j++){
                if (i%j == 0){
                    is = true;
                    break;
                }
            }

            if (!is){
                arr[k++] = i;
            }
        }
//
//        for (i=0; i<k; i++){
//            out.print(arr[i]+ " ");
//        }
//        out.println();
//        out.println(k);
//        out.flush();


        sum = (n * (n+1) / 2);

        if (sum % 2 == 0){
            out.println("Yes\n" +  1 + " " + 2);
            out.print(n-1+ " ");
            for (i=1; i<=n; i++){
                if (i == 2)
                    continue;
                else out.print(i + " ");
            }

        }
        else {

            for (i=0; i<k && arr[i] <= n; i++){
                if (sum % arr[i] == 0){


                    j = i;

                    out.println("Yes\n" + 1 + " " + arr[i]);
                    out.print(n-1+ " ");

                    for (i=1; i<=n; i++){

                        if (i == arr[j])
                            continue;

                        else {
                            if (i != n)
                            out.print(i + " ");
                            else out.print(i);
                        }

                    }


                    out.flush();
                    return;

                }
            }
            out.println("No");



        }

        //mergeSort(arr, 0, 7);


        out.flush();


    }



    static void  mergeSort(int arr[], int l, int r)
    {
        if (l < r)
        {
            // Same as (l+r)/2, but avoids overflow for
            // large l and h
            int m = l+(r-l)/2;

            // Sort first and second halves
            mergeSort(arr, l, m);
            mergeSort(arr, m+1, r);

            merge(arr, l, m, r);
        }
    }


    static void merge(int arr[], int l, int m, int r)
    {
        int i, j, k;
        int n1 = m - l + 1;
        int n2 =  r - m;

        /* create temp arrays */
        int L[] = new int[n1],  R[] = new int[n2];

        /* Copy data to temp arrays L[] and R[] */
        for (i = 0; i < n1; i++)
            L[i] = arr[l + i];
        for (j = 0; j < n2; j++)
            R[j] = arr[m + 1+ j];

        /* Merge the temp arrays back into arr[l..r]*/
        i = 0; // Initial index of first subarray
        j = 0; // Initial index of second subarray
        k = l; // Initial index of merged subarray
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }

    /* Copy the remaining elements of L[], if there
       are any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }

    /* Copy the remaining elements of R[], if there
       are any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }






    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

}