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 PYTHON):

Solution:
from __future__ import division, print_function
import threading,bisect,math,heapq,sys
# threading.stack_size(2**27)
# sys.setrecursionlimit(10**4)
from sys import stdin, stdout
i_m=9223372036854775807    
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input()) 
prime=[]
def dfs(n,d,v,a,c):
    v[n]=1
    if n in d:
        x=d[n]
    else:
        a[n]+=c
        x=[]
    p=a[n]
    for i in x:
        if i not in v:
            dfs(i,d,v,a,c)
            p+=a[i]
    a[n]=p
    return p


"""**************************MAIN*****************************"""
def main():
    t=inin()
    for _ in range(t):
        n=inin()
        a=ain()
        ans=[]
        for i in range(0,n,2):
            ans.append((1,i+1,i+2))
            # a[i]=a[i]+a[i+1]
            ans.append((2,i+1,i+2))
            # a[i+1]=a[i+1]-a[i]
            ans.append((2,i+1,i+2))
            # a[i+1]=a[i+1]-a[i]
            ans.append((1,i+1,i+2))
            # a[i]=a[i]+a[i+1]
            ans.append((2,i+1,i+2))
            # a[i+1]=a[i+1]-a[i]
            ans.append((2,i+1,i+2))
            # a[i+1]=a[i+1]-a[i]
        print(len(ans))
        for i in ans:
            print(*i)
            

        
        

        
"""*******************************************************"""

######## Python 2 and 3 footer by Pajenegod and c1729
fac=[]
def fact(n,mod):
    global fac
    fac.append(1)
    for i in range(1,n+1):
        fac.append((fac[i-1]*i)%mod)
    f=fac[:]
    return f
def nCr(n,r,mod):
    global fac
    x=fac[n]
    y=fac[n-r]
    z=fac[r]
    x=moddiv(x,y,mod)
    return moddiv(x,z,mod)
def moddiv(m,n,p):
    x=pow(n,p-2,p)
    return (m*x)%p
def GCD(x, y): 
    x=abs(x)
    y=abs(y)
    if(min(x,y)==0):
        return max(x,y)
    while(y): 
        x, y = y, x % y 
    return x 
def Divisors(n) : 
    l = []  
    ll=[]
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                ll.append(n//i)
    l.extend(ll[::-1])
    return l
def SieveOfEratosthenes(n): 
    global prime
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    f=[]
    for p in range(2, n): 
        if prime[p]: 
            f.append(p)
    return f
def primeFactors(n): 
    a=[]
    while n % 2 == 0: 
        a.append(2) 
        n = n // 2 
    for i in range(3,int(math.sqrt(n))+1,2):  
        while n % i== 0: 
            a.append(i) 
            n = n // i  
    if n > 2: 
        a.append(n)
    return a
"""*******************************************************"""
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
import os
from io import IOBase, BytesIO
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n') 
# Cout implemented in Python
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'
 
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'R' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'R' [0]:
        A.append(sign*numb)
    return A
 
# threading.Thread(target=main).start()
if __name__== "__main__":
  main()