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:
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#include <bits/stdc++.h>
using namespace std;
//---------------------------------//
void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}


void hat(){
	//tcT> constexpr bool needs_output_v = !is_printable_v<T> && is_iterable_v<T>;

	// ts: string representation to print
	long long a[0];
long long sum[0];
long long dp[0];
long long q[0];
long long minn[0];

    int n,c;
    scanf("%d%d",&n,&c);
    for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
    for(int i=1;i<=c;i++) sum[i]=sum[i-1]+a[i];
    for(int i=c+1;i<=n;i++) sum[i]=sum[i-1]+a[i]-a[i-c];
    for(int i=1;i<c;i++) dp[i]=dp[i-1]+a[i];
    q[1]=1;
    int l=1,r=1;
    for(int i=1;i<=n;i++){
    	while(l<=r&&q[l]<=i-c) l++;
    	while(l<=r&&a[q[r]]>a[i]) r--;
    	q[++r]=i,minn[i]=a[q[l]];
	}
    for(int i=c;i<=n;i++){
        dp[i]=min(dp[i-1]+a[i],dp[i-c]+sum[i]-minn[i]);
    }
    printf("%lld",dp[n]);
}

int ye_nahai(int x, char ch)
{
    int n;
    cin>>n;
    int arr[n];

     int max_ending_here = 0, max_so_far = INT_MIN;
    for (int i = 0; i < n; i++) {

        // include current element to previous subarray only
        // when it can add to a bigger number than itself.
        if (arr[i] <= max_ending_here + arr[i]) {
            max_ending_here += arr[i];
        }

        // Else start the max subarry from current element
        else {
            max_ending_here = arr[i];
        }
        if (max_ending_here > max_so_far)
            max_so_far = max_ending_here;
    }

    n *= 2;
    vector<int> a(n);
    for (int i = 0; i < n; ++i)
    {
    	cin>>a[i];
    }
    //sort(all(a));
    for(int i = 0; i < n/2; i++){
    	cout<<a[i]<<" "<<a[i+n/2]<<" ";
    }
    cout<<endl;
    cin>>n;
    n *= 2;
//    vector<int> a(n);
    for (int i = 0; i < n; ++i)
    {
    	cin>>a[i];
    }
    //sort(all(a));
    for(int i = 0; i < n/2; i++){
    	cout<<a[i]<<" "<<a[i+n/2]<<" ";
    }
    cout<<endl;
    cin>>n;
    n *= 2;
   // vector<int> a(n);
    for (int i = 0; i < n; ++i)
    {
    	cin>>a[i];
    }
    //sort(all(a));
    for(int i = 0; i < n/2; i++){
    	cout<<a[i]<<" "<<a[i+n/2]<<" ";
    }
    cout<<endl;
    cin>>n;
    n *= 2;
    //vector<int> a(n);
    for (int i = 0; i < n; ++i)
    {
    	cin>>a[i];
    }
    //sort(all(a));
    for(int i = 0; i < n/2; i++){
    	cout<<a[i]<<" "<<a[i+n/2]<<" ";
    }
    cout<<endl;
    cin>>n;
    n *= 2;
    //vector<int> a(n);
    for (int i = 0; i < n; ++i)
    {
    	cin>>a[i];
    }
    //sort(all(a));
    for(int i = 0; i < n/2; i++){
    	cout<<a[i]<<" "<<a[i+n/2]<<" ";
    }
    cout<<endl;
    cin>>n;
    n *= 2;
    //vector<int> a(n);
    for (int i = 0; i < n; ++i)
    {
    	cin>>a[i];
    }
    //sort(all(a));
    for(int i = 0; i < n/2; i++){
    	cout<<a[i]<<" "<<a[i+n/2]<<" ";
    }
    cout<<endl;
    cin>>n;
    n *= 2;
    //vector<int> a(n);
    for (int i = 0; i < n; ++i)
    {
    	cin>>a[i];
    }
    //sort(all(a));
    for(int i = 0; i < n/2; i++){
    	cout<<a[i]<<" "<<a[i+n/2]<<" ";
    }
    cout<<endl;
    cin>>n;
    n *= 2;
    //vector<int> a(n);
    for (int i = 0; i < n; ++i)
    {
    	cin>>a[i];
    }
    //sort(all(a));
    for(int i = 0; i < n/2; i++){
    	cout<<a[i]<<" "<<a[i+n/2]<<" ";
    }
    cout<<endl;
}



#ifndef ONLINE_JUDGE
#define deb(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define deb(x...)
#endif
//------------------------------//

typedef long long int ll;
typedef pair<int,int> pi;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<pi> vpi;
typedef vector<pll> vpll;

#define  IOS          ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define  all(x)       (x).begin(), (x).end()
#define  ps(x,y)      fixed<<setprecision(y)<<x
#define  rep(i,a,b)   for(ll i=a ; i<b ; ++i)

const ll MOD = 1e9 + 7;
const ll INF = 1e18;
const ll maxn = 1e6 + 4;

void solve(){
int n;
    cin>>n;

    int arr[n];
    for(int i=0; i<n; i++)
        cin>>arr[i];

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

    for(int i=1; i<=n; i+=2)
    {
        cout<<1<<" "<<i<<" "<<i+1<<endl;
        cout<<1<<" "<<i<<" "<<i+1<<endl;
        cout<<2<<" "<<i<<" "<<i+1<<endl;
        cout<<1<<" "<<i<<" "<<i+1<<endl;
        cout<<1<<" "<<i<<" "<<i+1<<endl;
        cout<<2<<" "<<i<<" "<<i+1<<endl;
    }

}

int main(){
  IOS
	int test;
	cin>>test;

	while(test--)
  		solve();
}

// integer overflow
// var vs loop var