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.*;
 
public class Main{
    public static void main(String[] args) throws IOException{
    	In in = new In();
    	int n = in.pint();
    	int[] a = new int[n + 1];
    	in.tok();
    	for(int i = 1; i <= n; i++) {
    		a[i] = in.pint();
    	}
    	boolean f = n % 2 == 0;
    	if(n % 2 == 0) {n--;}
    	StringBuilder sb = new StringBuilder();
    	int kp = 0;
    	for(int i = 1; i + 2 <= n; i += 2) {
    		sb.append(i);
    		sb.append(' ');
    		sb.append(i + 1);
    		sb.append(' ');
    		sb.append(i + 2);
    		sb.append('\n');
    		kp++;
    		int t = a[i] ^ a[i + 1] ^ a[i + 2];
    		a[i] = t; a[i + 1] = t; a[i + 2] = t;
    	}
    	for(int i = n - 1; i > 0; i -= 2) {
    		sb.append(i);
    		sb.append(' ');
    		sb.append(i - 1);
    		sb.append(' ');
    		sb.append(n);
    		sb.append('\n');
    		kp++;
    		int t = a[i] ^ a[i - 1] ^ a[n];
    		if(a[i] != a[i - 1]) {System.out.println("?");}
    		a[i] = t;
    		a[i - 1] = t;
    		a[n] = t;
    	}
    	if(!f || a[n] == a[n + 1]) {
    		System.out.println("YES");
    		System.out.println(kp);
    		System.out.print(sb);
    	}else {
    		System.out.println("NO");
    	}
    }
}   
class In{
    BufferedReader in;
    StringTokenizer st = new StringTokenizer("");
    public In(){
        in = new BufferedReader(new InputStreamReader(System.in));
    }
    
    int pint() throws IOException{
        if(st.hasMoreTokens()) {return Integer.parseInt(st.nextToken());}
        else {return Integer.parseInt(in.readLine());}
    } 
    double pdbl() throws IOException{
        if(st.hasMoreTokens()) {return Double.parseDouble(st.nextToken());}
        else {return Double.parseDouble(in.readLine());}
    }
    long plng() throws IOException{
        if(st.hasMoreTokens()) {return Long.parseLong(st.nextToken());}
        else {return Long.parseLong(in.readLine());}
    }
    char pchr() throws IOException{
        if(st.hasMoreTokens()) {return st.nextToken().charAt(0);}
        else {return in.readLine().charAt(0);}
    }
    String pstr() throws IOException{
        if(st.hasMoreTokens()) {return st.nextToken();}
        else {return in.readLine();}
    }
    String readLine() throws IOException{
        return in.readLine();
    }
    boolean ready() throws IOException {return in.ready();}
    boolean readyN() throws IOException{return in.ready() || st.hasMoreTokens();}
    boolean tokens() {return st.hasMoreTokens();}
    void tok() throws IOException{st = new StringTokenizer(in.readLine());}
    void skip() throws IOException{in.readLine();}
}