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]. 
/*
Author: Racer5x
 ***************************************   UNAUTHORISED COPYING OF CODE PROHIBITED   **********************************
*/
 
// #pragma GCC optimize("O3")
// #pragma GCC optimize("Ofast")
// #pragma GCC optimize("unroll-loops")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
 
#include<bits/stdc++.h>
#define int         long long
#define double      long double
#define pb          emplace_back
#define pf          emplace_front
#define pii         pair<int,int>
#define vi          vector<int>
#define vii         vector<pii>
#define mi          map<int,int>
#define mii         map<pii,int>
#define all(a)      (a).begin(),(a).end()
#define rall(a)     (a).rbegin(),(a).rend()
#define x           first
#define y           second
#define endl        '\n'
#define hell        998244353 
#define PI          3.141592653589
#define tezz        ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define MAX         2000000000000000000
#define M           1000000007
using namespace std;
 
signed main()
{
    tezz

    int t;
    cin>>t;

    while(t--){
        int n;
        cin>>n;

        vi v(n + 1);
        for(int i=0; i<=n; i++) v[i] = i;

        vii ans;

        if(n <= 32){
        	for(int i=n - 1; i>2; i--) ans.pb(i, n), v[i] = 1;

        	while(v[n] > 1){
        		ans.pb(n, 2);
        		v[n] = (v[n] + 1)/2;
        	}
        }
        else{
        	for(int i=n - 1; i>4; i--) if(i != 32) ans.pb(i, n), v[i] = 1;

        	while(v[n] > 1){
        		ans.pb(n, 32);
        		v[n] = (v[n] + 31)/32;
        	}

        	while(v[32] > 1){
        		ans.pb(32, 4);
        		v[32] = (v[32] + 3) / 4;
        	}

        	ans.pb(3, 4);
        	ans.pb(4, 2);
        	ans.pb(4, 2);
        }

        // for(int i=1; i<=n; i++) cout<<v[i]<<' ';
        // 	cout<<endl;

        cout<<ans.size()<<endl;
        for(auto it: ans) cout<<it.x<<' '<<it.y<<endl;
    }
}