Problem: Ksenia has an array a consisting of n positive integers a_1, a_2, …, a_n. 

In one operation she can do the following: 

  * choose three distinct indices i, j, k, and then 
  * change all of a_i, a_j, a_k to a_i ⊕ a_j ⊕ a_k simultaneously, where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). 



She wants to make all a_i equal in at most n operations, or to determine that it is impossible to do so. She wouldn't ask for your help, but please, help her!

Input

The first line contains one integer n (3 ≤ n ≤ 10^5) — the length of a.

The second line contains n integers, a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — elements of a.

Output

Print YES or NO in the first line depending on whether it is possible to make all elements equal in at most n operations.

If it is possible, print an integer m (0 ≤ m ≤ n), which denotes the number of operations you do.

In each of the next m lines, print three distinct integers i, j, k, representing one operation. 

If there are many such operation sequences possible, print any. Note that you do not have to minimize the number of operations.

Examples

Input


5
4 2 1 7 2


Output


YES
1
1 3 4

Input


4
10 4 49 22


Output


NO

Note

In the first example, the array becomes [4 ⊕ 1 ⊕ 7, 2, 4 ⊕ 1 ⊕ 7, 4 ⊕ 1 ⊕ 7, 2] = [2, 2, 2, 2, 2].

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
import java.math.*;
 
public class D
{
    public static void process(int test_number)throws IOException
    {
        int n = ni(), arr[] = new int[n + 1], xorSum = 0;
        for(int i = 1; i <= n; i++){
            arr[i] = ni();     
            xorSum ^= arr[i];
        }

        if(n % 2 == 0){
            if(xorSum != 0){
                pn("NO");
                return ;
            }
            else
                n--;
        }

        pn("YES" + "\n" + (n - 1));
        for(int i = 1; i < n; i += 2)
            pn(i + " " + (i + 1) +  " " + n);

        for(int i = 1; i < n; i += 2)
            pn(i + " " + (i + 1) +  " " + n);
    }
 
    static final long mod = (long)1e9+7l;
    static boolean DEBUG = true;
    static FastReader sc;
    static PrintWriter out;
    public static void main(String[]args)throws IOException
    {
        out = new PrintWriter(System.out);
        sc = new FastReader();
 
        long s = System.currentTimeMillis();
        int t = 1;
        //t = ni();
        for(int i = 1; i <= t; i++)
            process(i);
 
        out.flush();
        System.err.println(System.currentTimeMillis()-s+"ms");
    }

    static void trace(Object... o){ if(!DEBUG) return; System.err.println(Arrays.deepToString(o)); };    
    static void pn(Object o){ out.println(o); }
    static void p(Object o){ out.print(o); }
    static int ni()throws IOException{ return Integer.parseInt(sc.next()); }
    static long nl()throws IOException{ return Long.parseLong(sc.next()); }
    static double nd()throws IOException{ return Double.parseDouble(sc.next()); }
    static String nln()throws IOException{ return sc.nextLine(); }
    static long gcd(long a, long b){ return (b==0)?a:gcd(b,a%b);}
    static int gcd(int a, int b){ return (b==0)?a:gcd(b,a%b); }
    
    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(); 
        } 
  
        String nextLine(){ 
            String str = ""; 
            try{ str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } 
            return str; 
        } 
    } 
}