Problem: <image>

While trading on his favorite exchange trader William realized that he found a vulnerability. Using this vulnerability he could change the values of certain internal variables to his advantage. To play around he decided to change the values of all internal variables from a_1, a_2, …, a_n to -a_1, -a_2, …, -a_n. For some unknown reason, the number of service variables is always an even number.

William understands that with his every action he attracts more and more attention from the exchange's security team, so the number of his actions must not exceed 5 000 and after every operation no variable can have an absolute value greater than 10^{18}. William can perform actions of two types for two chosen variables with indices i and j, where i < j:

  1. Perform assignment a_i = a_i + a_j 
  2. Perform assignment a_j = a_j - a_i 

William wants you to develop a strategy that will get all the internal variables to the desired values.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 20). Description of the test cases follows.

The first line of each test case contains a single even integer n (2 ≤ n ≤ 10^3), which is the number of internal variables.

The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9), which are initial values of internal variables.

Output

For each test case print the answer in the following format:

The first line of output must contain the total number of actions k, which the strategy will perform. Note that you do not have to minimize k. The inequality k ≤ 5 000 must be satisfied. 

Each of the next k lines must contain actions formatted as "type i j", where "type" is equal to "1" if the strategy needs to perform an assignment of the first type and "2" if the strategy needs to perform an assignment of the second type. Note that i < j should hold.

We can show that an answer always exists.

Example

Input


2
4
1 1 1 1
4
4 3 1 2


Output


8
2 1 2
2 1 2
2 1 3
2 1 3
2 1 4
2 1 4
1 1 2
1 1 2
8
2 1 4
1 2 4
1 2 4
1 2 4
1 3 4
1 1 2
1 1 2
1 1 4

Note

For the first sample test case one possible sequence of operations is as follows:

  1. "2 1 2". Values of variables after performing the operation: [1, 0, 1, 1] 
  2. "2 1 2". Values of variables after performing the operation: [1, -1, 1, 1] 
  3. "2 1 3". Values of variables after performing the operation: [1, -1, 0, 1] 
  4. "2 1 3". Values of variables after performing the operation: [1, -1, -1, 1] 
  5. "2 1 4". Values of variables after performing the operation: [1, -1, -1, 0] 
  6. "2 1 4". Values of variables after performing the operation: [1, -1, -1, -1] 
  7. "1 1 2". Values of variables after performing the operation: [0, -1, -1, -1] 
  8. "1 1 2". Values of variables after performing the operation: [-1, -1, -1, -1] 



For the second sample test case one possible sequence of operations is as follows:

  1. "2 1 4". Values of variables after performing the operation: [4, 3, 1, -2] 
  2. "1 2 4". Values of variables after performing the operation: [4, 1, 1, -2] 
  3. "1 2 4". Values of variables after performing the operation: [4, -1, 1, -2] 
  4. "1 2 4". Values of variables after performing the operation: [4, -3, 1, -2] 
  5. "1 3 4". Values of variables after performing the operation: [4, -3, -1, -2] 
  6. "1 1 2". Values of variables after performing the operation: [1, -3, -1, -2] 
  7. "1 1 2". Values of variables after performing the operation: [-2, -3, -1, -2] 
  8. "1 1 4". Values of variables after performing the operation: [-4, -3, -1, -2] 

Solution (in JAVA):

Solution:
import java.util.*;
import java.lang.*;
import java.io.*;
import java.awt.*;

/*AUTHOR - SHIVAM GUPTA */

// U KNOW THAT IF THIS DAY WILL BE URS THEN NO ONE CAN DEFEAT U HERE ,JUst keep faith in ur strengths .................................................. 
// ASCII = 48 + i ;// 2^28 = 268,435,456  > 2* 10^8 // log 10 base 2 = 3.3219 
// odd:: (x^2+1)/2 , (x^2-1)/2 ; x>=3// even:: (x^2/4)+1 ,(x^2/4)-1  x >=4 
// FOR ANY ODD NO N : N,N-1,N-2,ALL ARE PAIRWISE COPRIME,THEIR COMBINED LCM IS PRODUCT OF ALL THESE NOS
// two consecutive odds are always coprime to each other, two consecutive even have always gcd  = 2 ;

// Rectangle r = new Rectangle(int x , int y,int widht,int height) 
//Creates a rect. with bottom left cordinates as (x, y) and top right as ((x+width),(y+height))

//BY DEFAULT Priority Queue is MIN in nature in java
//to use as max , just push with negative sign and change sign after removal 

// We can make a sieve of max size 1e7 .(no time or space issue) 
// In 1e7 starting nos we have about 66*1e4 prime nos
// In 1e6 starting nos we have about 78,498 prime nos

 public class Main
 //public class Solution
{
    static PrintWriter out;
    
	static class FastReader{
		BufferedReader br;
		StringTokenizer st;
		public FastReader(){
			br=new BufferedReader(new InputStreamReader(System.in));
			out=new PrintWriter(System.out);
		}
		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;
		}
	}
	
///////////////////////////////////////////////////////////////////////////////////////////////////////
 public static int sumOfDigits(long n)
 {
  if( n< 0)return -1 ;
  int sum = 0;
  while( n > 0){ sum = sum + (int)( n %10) ; n /= 10 ; } 
  return sum ;  
 }
 
//////////////////////////////////////////////////////////////////////////////////////////////////////
public static void swapArray(int[] arr , int start , int end)
{
    while(start<end){ int temp = arr[start] ;arr[start] = arr[end]; arr[end] = temp;  start++ ;end-- ;}
   
}
//////////////////////////////////////////////////////////////////////////////////

static long factorial(long a)
{
    if(a== 0L || a==1L)return 1L ;
   return a*factorial(a-1L) ;
}
///////////////////////////////////////////////////////////////////////////////
public static int[][] rotate(int[][] input){

int n =input.length;int m = input[0].length ;
int[][] output = new int [m][n];
for (int i=0; i<n; i++)
	for (int j=0;j<m; j++)
		output [j][n-1-i] = input[i][j];
return output;
}
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// ////////////////////////////////////////////////   
public static String reverse(String input)
{
  StringBuilder str  = new StringBuilder("") ;
   
    for(int i =input.length()-1 ; i >= 0  ; i-- )
    {
        str.append(input.charAt(i));
    }
    
return str.toString() ;
}
///////////////////////////////////////////////////////////////////////////////////////////

public static int xorOfFirstN(int n)
{
  if( n % 4 ==0)return n ; else if( n % 4 == 1)return 1 ;
  else if( n % 4 == 2)return n+1 ; else return 0 ;
 }
//////////////////////////////////////////////////////////////////////////////////////////////

public static int gcd(int a, int b )
{
   if(b==0)return a ;else return gcd(b,a%b) ; 
}

public static long gcd(long a, long b )
{
if(b==0)return a ;else return gcd(b,a%b) ; 
}
///////////////////////////////////////////////////////////////////////////////////////////

public static int lcm(int a, int b ,int c )
{
int temp = lcm(a,b) ;int ans =  lcm(temp ,c) ;return ans  ;
}

////////////////////////////////////////////////////////////////////////////////////////
    
public static int lcm(int a , int b )
{
   int gc = gcd(a,b);return (a/gc)*b ;
}


public static long lcm(long a , long b )
{
    long gc = gcd(a,b);return (a/gc)*b;
}


///////////////////////////////////////////////////////////////////////////////////////////
static boolean isPrime(long n)
{ 
   if(n==1)return false  ;  
    for(long i = 2L; i*i <= n ;i++){ if(n% i ==0){return false; } }
    return true ;
} 

static boolean isPrime(int n)
{
    if(n==1)return false  ;  
    for(int i = 2; i*i <= n ;i++){ if(n% i ==0){return false ; } }
    return true ;
}      
////////////////////////////////////////////////////////////////////////////////////////////////////////

///////////////////////////////////////////////////////////////////////////

static int sieve =  1000000 ;

 
static boolean[] prime =  new boolean[sieve + 1] ;

public static void sieveOfEratosthenes() 
    { 
        // FALSE == prime and 1 // TRUE ==  COMPOSITE
        
        // time complexity = 0(NlogLogN)== o(N)
        
        // gives prime nos bw 1 to N // size - 1e7(at max)
        
        for(int i = 4; i<= sieve ; i++)
        {
            prime[i] = true  ; i++ ;
           
        }
        
        for(int p = 3; p*p <= sieve; p++) 
        { 
           if(prime[p] == false) 
            { 
                    for(int i = p*p; i <= sieve; i += p) 
                    prime[i] = true; 
            } 
            
            p++ ;
        } 
          
    } 
 
///////////////////////////////////////////////////////////////////////////////////
public static void sortD(int[] arr , int s  , int e)
{
      sort(arr ,s , e) ;
      int i =s ; int j = e  ;
      while( i < j){  int temp = arr[i] ;arr[i] =arr[j] ; arr[j] = temp ; i++ ; j-- ;}
}

/////////////////////////////////////////////////////////////////////////////////////////

public static long countSubarraysSumToK(long[] arr ,long sum )
    {
      HashMap<Long,Long> map = new HashMap<>() ;
       
      int n = arr.length ; long prefixsum = 0 ; long count = 0L ;
        for(int i  = 0; i < n ; i++)
      {
          prefixsum  = prefixsum +  arr[i] ;
           
          if(sum == prefixsum)count = count+1 ;
           
          if(map.containsKey(prefixsum -sum))count = count + map.get(prefixsum -sum) ;
         
          if(map.containsKey(prefixsum )) map.put(prefixsum , map.get(prefixsum) +1 );
          else map.put(prefixsum , 1L );
      }
        
      return count  ;  
    }

///////////////////////////////////////////////////////////////////////////////////////////////


// KMP ALGORITHM : TIME COMPL:O(N+M) 
// FINDS THE OCCURENCES OF PATTERN AS A  SUBSTRING IN STRING
//RETURN THE ARRAYLIST OF INDEXES 
// IF SIZE OF LIST IS ZERO MEANS PATTERN IS NOT PRESENT IN STRING


public static ArrayList<Integer> kmpAlgorithm(String str , String pat)
     {
        ArrayList<Integer> list =new ArrayList<>();
        
        int n = str.length() ;
        int m = pat.length() ;
        
        String q = pat + "#" + str ;
        
        int[] lps  =new int[n+m+1] ;
        
         longestPefixSuffix(lps, q,(n+m+1)) ;
         
         
         for(int i =m+1 ; i < (n+m+1) ; i++ )
         {
             if(lps[i] == m)
             {
                 list.add(i-2*m) ;
             }
         }
        
        return list ; 
         
        
    }
    

public static void longestPefixSuffix(int[] lps ,String str , int n)
    {
        lps[0] =  0 ;
        
        for(int i = 1  ; i<= n-1; i++)
        {
          int l = lps[i-1] ;
           
          while( l > 0 && str.charAt(i) != str.charAt(l))
          {
              l = lps[l-1] ;
          }
           
          if(str.charAt(i) == str.charAt(l))
          {l++ ;}
              
              lps[i] = l ; 
        }
        
    }
    
 ///////////////////////////////////////////////////////////////////////////////////////////////////

 // CALCULATE TOTIENT Fn FOR ALL VALUES FROM 1 TO n
    // TOTIENT(N) = count of nos less than n and grater than 1 whose gcd with n is 1 
    // or n and the no  will be coprime in nature
    //time : O(n*(log(logn)))
    
    public static void eulerTotientFunction(int[] arr ,int n )
    {
      for(int i = 1; i <= n  ;i++)arr[i] =i  ;
      
       for(int i= 2 ; i<= n ;i++)
      {
          if(arr[i] == i)
          {
              arr[i] =i-1 ;
              
              for(int j =2*i ; j<= n  ; j+= i )
              {
                  arr[j] = (arr[j]*(i-1))/i ;
              }
              
          }
      }
        
    }
	
/////////////////////////////////////////////////////////////////////////////////////////////
public static long nCr(int n,int k)
{
    long ans=1L;
    k=k>n-k?n-k:k;
    int j=1;
    for(;j<=k;j++,n--)
    {
        if(n%j==0)
        {
            ans*=n/j;
        }else
        if(ans%j==0)
        {
            ans=ans/j*n;
        }else
        {
            ans=(ans*n)/j;
        }
    }
    return ans;
}

///////////////////////////////////////////////////////////////////////////////////////////

public static ArrayList<Integer> allFactors(int n)
{   
      ArrayList<Integer> list = new ArrayList<>() ;
      
    for(int i = 1; i*i <= n ;i++)
    {
          if( n % i == 0)
          {
              if(i*i == n)list.add(i) ;
              else{ list.add(i) ;list.add(n/i) ; }
          }
    }
      
     return list ; 
}


public static ArrayList<Long> allFactors(long n)
{   
      ArrayList<Long> list = new ArrayList<>() ;
      
    for(long i = 1L; i*i <= n ;i++)
    {
          if( n % i == 0)
          {
              if(i*i == n) list.add(i) ;
              else{  list.add(i) ; list.add(n/i) ; }
             
          }
    }
      
     return list ; 
}
////////////////////////////////////////////////////////////////////////////////////////////////////

  static final int MAXN = 1000001; 
       
    
    static int spf[] = new int[MAXN]; 
   
    static void sieve() 
    { 
       for (int i=1; i<MAXN; i++) spf[i] = i; // ALL NOS HAVE SPF[i] =i AT START
       
        for (int i=4; i<MAXN; i+=2) //ALL EVEN NOS HAVE SPF[i] = 2 
            spf[i] = 2; 
       
        for (int i=3; i*i<MAXN; i++) 
        { 
           if (spf[i] == i) 
            { 
               for (int j=i*i; j<MAXN; j+=i) 
       
                    if (spf[j]==j) spf[j] = i;  
            } 
        } 
    } 
       

    static ArrayList<Integer> getPrimeFactorization(int x) 
    { 
        ArrayList<Integer> ret = new ArrayList<Integer>(); 
        while (x != 1) 
        { 
            ret.add(spf[x]); 
            x = x / spf[x]; 
        } 
        return ret; 
    } 
       
 //////////////////////////////////////////////////////////////////////////////////////////////////
 //////////////////////////////////////////////////////////////////////////////////////////////////
   
public static void merge(int arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;
 
        /* Create temp arrays */
        int L[] = new int[n1];
        int R[] = new int[n2];
 
       //Copy data to temp arrays
        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];
 
 
        /* Merge the temp arrays */
 
        // Initial indexes of first and second subarrays
        int i = 0, j = 0;
 
        // Initial index of merged subarry array
        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++;
        }
 
        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }
 
        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
 
    // Main function that sorts arr[l..r] using
    // merge()
  public static void sort(int arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;
 
            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);
 
            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }

public static void sort(long arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;
 
            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);
 
            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }


public static void merge(long arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;
 
        /* Create temp arrays */
        long L[] = new long[n1];
        long R[] = new long[n2];
 
        //Copy data to temp arrays
        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];
 
 
        /* Merge the temp arrays */
 
        // Initial indexes of first and second subarrays
        int i = 0, j = 0;
 
        // Initial index of merged subarry array
        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++;
        }
 
        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }
 
        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
 

 /////////////////////////////////////////////////////////////////////////////////////////
 
	public static long knapsack(int[] weight,long value[],int maxWeight){

        int n=  value.length ;

	//dp[i] stores the profit with KnapSack capacity "i" 
  long []dp = new long[maxWeight+1]; 
      
    //initially profit with 0 to W KnapSack capacity is 0 
    Arrays.fill(dp, 0); 
  
    // iterate through all items 
    for(int i=0; i < n; i++)  
      
        //traverse dp array from right to left 
        for(int j = maxWeight; j >= weight[i]; j--) 
            dp[j] = Math.max(dp[j] , value[i] + dp[j - weight[i]]); 
              
    /*above line finds out maximum of dp[j](excluding ith element value) 
    and val[i] + dp[j-wt[i]] (including ith element value and the 
    profit with "KnapSack capacity - ith element weight") */
    return dp[maxWeight]; 
	}

///////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////


// to return max sum of any subarray in given array
public static long kadanesAlgorithm(long[] arr)
{
    if(arr.length == 0)return  0 ;    
    
    long[] dp = new long[arr.length] ; dp[0] = arr[0] ;long max =  dp[0] ;
      
    for(int i = 1; i <  arr.length ; i++)
      {
            if(dp[i-1] > 0) dp[i] = dp[i-1] + arr[i] ; else  dp[i] = arr[i] ;
            if(dp[i] >  max)max = dp[i] ;
    }
       return max  ;
      
}
/////////////////////////////////////////////////////////////////////////////////////////////
public static long kadanesAlgorithm(int[] arr)
{    if(arr.length == 0)return  0 ;
    
      long[] dp = new long[arr.length] ;dp[0] = arr[0] ; long max =  dp[0] ;
      for(int i = 1; i <  arr.length ; i++)
      {
            if(dp[i-1] > 0) dp[i] = dp[i-1] + arr[i] ;  else  dp[i] = arr[i] ;
           if(dp[i] >  max)max = dp[i] ;
      }
      
      return max  ;
}

/////////////////////////////////////////////////////////////////////////////////////////////////////////
    public static long power(int a  ,int b)
    {
        //time comp : o(logn) 
        long x = (long)(a) ; long n = (long)(b) ;
        if(n==0)return 1 ;if(n==1)return x;
         long ans =1L  ;
 
      while(n>0)
      {
          if(n % 2 ==1)ans = ans *x ;
         n = n/2L ;x =  x*x ; 
      }    
      return ans ;
    }
    
    public static long power(long a  ,long b)
    {
        //time comp : o(logn) 
        long x = (a) ;long n = (b) ;
         if(n==0)return 1L ;if(n==1)return x;
         long ans =1L  ;
        
      while(n>0)
      {
          if(n % 2 ==1)ans = ans *x ;
        
          n = n/2L ; x =  x*x ;
      }
       
      return ans ;
    }
    
//////////////////////////////////////////////////////////////////////////////////

/*
lowerBound - finds largest element equal or less than value paased
upperBound - finds smallest element equal or more than value passed
if not present return -1;
*/

public static long lowerBound(long[] arr,long k)
	{
		long ans=-1;
		
		int start=0;
		int end=arr.length-1;
		
		while(start<=end)
		{
			int mid=(start+end)/2;
			
			if(arr[mid]<=k)
			{
				ans=arr[mid];
				start=mid+1;
			}
			else
			{
				end=mid-1;
			}
			
		}
		
		return ans;
		
	}
	
	public static int lowerBound(int[] arr,int k)
	{
		int ans=-1;int start=0;int end=arr.length-1;
	
		while(start<=end)
		{
			int mid=(start+end)/2;
			
			if(arr[mid]<=k)
			{
				ans=arr[mid];
				start=mid+1;
			}
			else
			{
				end=mid-1;
			}
			
		}
		
		return ans;
		
	}
	
	
	public static long upperBound(long[] arr,long k)
	{
		long ans=-1;int start=0;int end=arr.length-1;
	
		while(start<=end)
		{
			int mid=(start+end)/2;
			
			if(arr[mid]>=k)
			{
				ans=arr[mid];
				end=mid-1;
			}
			else
			{
				start=mid+1;
			}
			
		}
		
		return ans;
	}
	
	
	public static int upperBound(int[] arr,int k)
	{
		int ans=-1; int start=0; 	int end=arr.length-1;

		while(start<=end)
		{
			int mid=(start+end)/2;
			
			if(arr[mid]>=k)
			{
				ans=arr[mid];
				end=mid-1;
			}
			else
			{
				start=mid+1;
			}
			
		}
		
		return ans;
	}
	
////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////

public static void printArray(int[] arr , int si ,int ei, int f)
{
    for(int i = si  ; i  <= ei ; i++){ out.print(arr[i] +" ") ; }  if(f==1)out.println() ;
}

public static void printArray(long[] arr , int si ,int ei, int f)
{
    for(int i = si  ; i  <= ei ; i++){ out.print(arr[i] +" ") ; }  if(f==1)out.println() ;
}


public static void printtwodArray(int[][] ans)
{
    for(int i = 0; i< ans.length ; i++)
    {
        for(int j  = 0 ; j <  ans[0].length ; j++)out.print(ans[i][j] +" "); out.println() ;
    }
    out.println() ;
}
/////////////////////////////////////////////////////////////////////////////////////////////////////
 ////////////////////////////////////////////////////////////////////////////////////////////////////
 
   static long modPow(long a, long x, long p) {
    //calculates a^x mod p in logarithmic time.
    
    a = a % p ;
    
    if(a == 0)return 0L ;
    if(x==0)return 1L;
    if(x==1)return a;
    
    long res = 1L;
    while(x > 0) {
        if( x % 2 != 0) {
            res = (res * a) % p;
        }
        a = (a * a) % p;
        x =x/2;
    }
    return res;
}
 
 
  static long modInverse(long a, long p) {
    //calculates the modular multiplicative of a mod p. //(assuming p is prime).
    return modPow(a, p-2, p);
}
 
 static long[] factorial =  new long[1000001] ;
 
 static void modfac(long mod)
 {
     factorial[0]=1L ; factorial[1]=1L ;
     
     for(int i = 2; i<= 1000000 ;i++)
     {
         factorial[i] = factorial[i-1] *(long)(i) ;
          factorial[i] = factorial[i] % mod ;
     }
    
     
 }
 
static long modBinomial(long n, long r, long p) {
// calculates C(n,r) mod p (assuming p is prime).
 
  if(n < r) return 0L ; 
 
    long num = factorial[(int)(n)] ;
    
    long den = (factorial[(int)(r)]*factorial[(int)(n-r)]) %  p ;
    
    
    long ans  = num*(modInverse(den,p)) ;
    
    ans = ans % p ;
    
    return ans  ;
 
    
}
 
 
 static void update(int val , long[] bit ,int n)
 {
     for( ; val <= n ; val += (val &(-val)) )
     {
         bit[val]++ ;
     }
    
     
 }
 
 
 static long query(int val , long[] bit , int n)
 {
     long ans = 0L; 
     for( ; val >=1 ; val-=(val&(-val))  )ans += bit[val];
     
     return ans ;
 }



 static int countSetBits(long n) 
    { 
        int count = 0; 
        while (n > 0) { 
            n = (n) & (n - 1L);  count++; 
            
        } 
        return count; 
    } 


static int abs(int x)
{ if(x<0)x=-x ; return x ; }
    
static long abs(long x)
{  if(x<0)x=-x ; return x ; }

////////////////////////////////////////////////////////////////////////////////////////////

 // calculate total no of nos greater than or equal to key in  sorted array arr
 
static  int bs(int[] arr, int s ,int e  ,int key)
{
        if( s> e)return 0 ;
        int mid = (s+e)/2 ;
        
        if(arr[mid] <key) return bs(arr ,mid+1,e , key) ;
        else  {return bs(arr ,s ,mid-1, key)  + e-mid+1;}
   
}
 
// static ArrayList<Integer>[] adj ;
// static int mod= 1000000007 ;


static class pair {
		int u;
	     int v;
 
	public	pair(int u,int v) {
			this.u = u ; this.v=v;
		
		}
	}
 
static class Dsu
{
    int n ;int[] par ; int[] size ;
    
    public Dsu(int n)
    {
        this.n = n ;
        par = new int[n] ;
        size = new int[n] ;
        
        for(int i = 0 ; i< n ;i++)
        {
            par[i] =i ;size[i] =1 ;
        }
    }
    
    // cheack if given element is root element of its set
    public boolean isRoot(int x)
    {
        return par[x] == x ;
    }
    
    // find root of given set
   public int findRoot(int x)
   {
       if(par[x] == x)
       {
          return par[x] ; 
       }
       
       par[x] = findRoot(par[x]) ; //path compression
       return par[x] ;
   }
   
   
   public boolean unionSet(int a, int b)
   {
       int rootA = findRoot(a);
       int rootB=findRoot(b) ;
       
       if(rootA==rootB)return true ;//they are alraedy present in same set/connected already(in graph)
       
       if(size[rootA] > size[rootB])
       {
           size[rootA] += size[rootB] ;
           par[rootB] =rootA ;
       }
       
       else{
           
           size[rootB] += size[rootA] ;
           par[rootA] =rootB ;
       }
       
       return false ;     // means we union both the sets now
   }
   
}

static void print(int t, int i , int j)
{
    out.println( (t)+" " + (i) +" " + (j) );
}

//////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////////////

public static void solve()
{
FastReader scn = new FastReader() ;


int testcase = 1;
 testcase = scn.nextInt() ;
for(int testcases =1  ; testcases <= testcase ;testcases++)
{
 
int n= scn.nextInt() ;
int[] arr = new int[n] ;

for(int i = 0 ; i  <  n;i++){
    arr[i] = scn.nextInt() ;
}


out.println(n*3) ;

for(int i =1 ; i <= n-1 ;i++)
{
    
     print(1,i,i+1) ;
     print(2,i,i+1) ;
      print(1,i,i+1) ;
       print(2,i,i+1) ;
        print(1,i,i+1) ;
         print(2,i,i+1) ;
    
    
    
    
    i++ ;
}




} // test case  end loop


out.flush() ;  
} // solve fn ends


public static void main (String[] args) throws java.lang.Exception
{     solve() ;    }
      
}
  
 class Pair 
{
  int first ;int second  ;
 
  public  Pair(int x, int y)
  {
      this.first = x ;this.second = y ;
  }
  
  @Override
 public boolean equals(Object obj)
 {
     if(obj == this)return true  ;
     
     if(obj == null)return false  ;
     
     if(this.getClass() != obj.getClass()) return false ;
     
     Pair other  = (Pair)(obj) ;
     
     if(this.first != other.first)return false  ;
     if(this.second != other.second)return false  ;
     
     return true  ;
     
 }
    
    @Override
    
    public int hashCode()
    {
        return this.first^this.second  ;
    }
      
@Override
	public String toString() {
	
	String ans = "" ;ans  += this.first ;	ans += " ";	ans += this.second ;
	return ans  ;
	}

}