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 CPP):

Solution:
#include <bits/stdc++.h>

const long long SZ = 1e5 + 7;
const long long inf = 1e18;
const long long mod = 1e9 + 7;
const long long MOD = 998244353;
long long opnmbr = 1;

#define ll                  long long
#define ld                  long double

#define pll                 pair<ll, ll>
#define vi                  vector<ll>
#define mi                  map<ll, ll>
#define umi                 unordered_map<ll, ll>

#define sz(x)               (ll)x.size()
#define endl                "\n"

#define pb                  push_back
#define F                   first
#define S                   second
#define mp(x,y)             make_pair(x,y)
#define all(a)              a.begin(),a.end()
#define rall(a)             a.rbegin(),a.rend()
#define rev(a)              reverse(all(a))
#define unq(a)              a.erase(std::unique(all(a)),a.end());

#define max(a,b)            ((a > b) ? a : b)
#define min(a,b)            ((a < b) ? a : b)

#define ci(X)               ll X; cin>>X
#define cii(X, Y)           ll X, Y; cin>>X>>Y
#define ciii(X, Y, Z)       ll X, Y, Z; cin>>X>>Y>>Z
#define ciiii(W, X, Y, Z)   ll W, X, Y, Z; cin>>W>>X>>Y>>Z

#define dbg(x)              cout<<#x<<"="<<(x)<<endl;
#define dbg2(x,y)           cout<<#x<<"="<<(x)<<" "<<#y<<"="<<(y)<<endl;
#define dbg3(x,y,z)         cout<<#x<<"="<<(x)<<" "<<#y<<"="<<(y)<<" "<<#z<<"="<<(z)<<endl;
#define dbg4(x,y,z,w)       cout<<#x<<"="<<(x)<<" "<<#y<<"="<<(y)<<" "<<#z<<"="<<(z)<<" "<<#w<<"="<<(w)<<endl;

#define kick()               cout<<"Case #"<<opnmbr++<<": ";
#define start_karo           ll ___t; cin>>___t; while (___t--)
#define chalo_achha_hai      ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define kardia               solve();

#define fr(i,a,b)           for(ll i=a;i<b;i++)
#define rf(i,a,b)           for(ll i=a;i>=b;i--)

#define mem0(X)             memset((X), 0, sizeof((X)))
#define mem1(X)             memset((X), -1, sizeof((X)))
#define setbits(x)          __builtin_popcountll(x)

#define yes()      cout<<"YES"<<endl
#define no()       cout<<"NO"<<endl
#define err()      cout<<"===========\n";
#define errA(A)    for(auto i:A) cout<<i<<" ";cout<<"\n";
#define err1(a)    cout<<#a<<" "<<a<<"\n"
#define err2(a,b)  cout<<#a<<" "<<a<<" "<<#b<<" "<<b<<"\n"

using namespace std;

ll powermod(ll a, ll b)
{
    ll ans = 1;
    while (b)
    {
        if (b & 1) ans = (ans * a) % MOD;
        b = b / 2;
        a = (a * a) % MOD;
    }
    return ans;
}

/* GCD and LCM of two numbers */

ll gcd(ll a, ll b)
{
    if (b == 0) return a;
    return gcd(b, a % b);
}

ll lcm(ll a, ll b)
{
    return (a / gcd(a, b)) * b;
}

/* Prime Numbers Sieve Method */

const long long nod = 1000000;
vector<char> is_prime(nod+1, true);

vector<int> prime;

void sieve(){

is_prime[0] = is_prime[1] = false;
for (int i = 2; i * i <= nod; i++) {
    if (is_prime[i]) {
        for (int j = i * i; j <= nod; j += i)
            is_prime[j] = false;
                     }
              }

     for(int i=0;i<=nod;i++)
     {
         if(is_prime[i])
           prime.push_back(i);

     }

 }

/*//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////*/
/*..................................................................................................................................................................................................*/
/*//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////*/

 
void solve()
{
    
      ci(n);

    fr(i,0,n)
     {
         int x;
         cin >>x;
     }


    cout<<n/2 * 6<<endl;

    for(int i=0;i<n;i+=2)
    {
          int turn = 1;

         for(int j=0;j<6;j++)
         {  
            if(turn)
              {cout<<1<<" "<<i+1<<" "<<i+2<<endl; turn = 0; }
            else
            {
                cout<<2<<" "<<i+1<<" "<<i+2<<endl; turn = 1;
            }
            
                
        }
    }


       
}

signed main()
{

  chalo_achha_hai

     #ifndef ONLINE_JUDGE
     freopen("input.txt" , "r" , stdin); freopen("output.txt" , "w" , stdout);
     #endif

     //sieve();
      start_karo
       {
          //kick()
            kardia
       }
   // kardia
  return 0;
}