Problem: You have an array a_1, a_2, ..., a_n where a_i = i.

In one step, you can choose two indices x and y (x ≠ y) and set a_x = \left⌈ (a_x)/(a_y) \right⌉ (ceiling function).

Your goal is to make array a consist of n - 1 ones and 1 two in no more than n + 5 steps. Note that you don't have to minimize the number of steps.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

The first and only line of each test case contains the single integer n (3 ≤ n ≤ 2 ⋅ 10^5) — the length of array a.

It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.

Output

For each test case, print the sequence of operations that will make a as n - 1 ones and 1 two in the following format: firstly, print one integer m (m ≤ n + 5) — the number of operations; next print m pairs of integers x and y (1 ≤ x, y ≤ n; x ≠ y) (x may be greater or less than y) — the indices of the corresponding operation.

It can be proven that for the given constraints it's always possible to find a correct sequence of operations.

Example

Input


2
3
4


Output


2
3 2
3 2
3
3 4
4 2
4 2

Note

In the first test case, you have array a = [1, 2, 3]. For example, you can do the following: 

  1. choose 3, 2: a_3 = \left⌈ (a_3)/(a_2) \right⌉ = 2 and array a = [1, 2, 2]; 
  2. choose 3, 2: a_3 = \left⌈ 2/2 \right⌉ = 1 and array a = [1, 2, 1]. 

You've got array with 2 ones and 1 two in 2 steps.

In the second test case, a = [1, 2, 3, 4]. For example, you can do the following: 

  1. choose 3, 4: a_3 = \left⌈ 3/4 \right⌉ = 1 and array a = [1, 2, 1, 4]; 
  2. choose 4, 2: a_4 = \left⌈ 4/2 \right⌉ = 2 and array a = [1, 2, 1, 2]; 
  3. choose 4, 2: a_4 = \left⌈ 2/2 \right⌉ = 1 and array a = [1, 2, 1, 1]. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll> pi;
ll mod=1000000007;

ll bpow(ll a, ll b)
{
    ll res=1;
    while(b>0)
    {
        if(b&1)
        {
            res=res*a;
            res%=mod;
        }
        a*=a;
        a%=mod;
        b>>=1;
    }
    return res;
}

void solve()
{
    ll n; cin>>n;
    if(n<8)
    {
        ll k=(n<=4 ? 2 : 3);
        cout<<k+n-3<<endl;
        for(ll i=n-1; i>=3; i--)
        {cout<<i<<" "<<n<<endl;}
        while(k--)
        {
            cout<<n<<" "<<2<<endl;
        }
    }
    else
    {
        ll k=8, c=1;
        while(k<n)
        {
            k*=8;
            c++;
        }
        cout<<(n-1)+c<<endl;
        for(ll i=n-1; i>=3; i--)
        {
            if(i!=8)
            {cout<<i<<" "<<n<<endl;}
        }
        if(c!=1)
        {
            while(c--)
            {
                cout<<n<<" "<<8<<endl;
            }
        }
        for(ll i=0; i<3; i++)
        {cout<<8<<" "<<2<<endl;}
        
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    ll t; cin>>t;
    while(t--)
    {
        solve();
    }
	// your code goes here
	return 0;
}