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 <iostream>
#include <vector>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <algorithm>
#include <sstream>
#include <iomanip>
#include <set>
#include <fstream>
#include <map>
#include <stack>
#include <string>
#include <cstring>
#include <cmath>
#include <cstdio>
#define ll long long
#define ld long double
#define INF 1000000
#define MOD 1000000007
using namespace std;
void solved() {
    ll n,k,elem;
    cin >> n;
    ll help1=ceil(sqrt(n));
    ll help2=n;
    vector<pair<ll,ll>> ans;
    while (help2>2){
        for (int i=help1+1;i<help2;i++){
            //cout << i << ' '<< i+1 << endl;
            ans.push_back({i,i+1});
        }
        //cout << help2 << ' ' << help1 << endl;
        //cout << help2 << ' ' << help1 << endl;
        ans.push_back({help2,help1});
        ans.push_back({help2,ceil((ld)help2/(ld)help1)});;
        help2=help1;
        help1=ceil(sqrt(help1));
    }
    cout << ans.size() << endl;
    for (auto v:ans){
        cout << v.first << ' ' << v.second << endl;
    }
}
int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cout.setf(ios::fixed);
    cout.precision(10);
    int t;
    cin >> t;
    //t=1;
    //ll t=100;
    while (t--) {
        solved();
    }
    return 0;
}