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;
struct chestie
{
    int alfa, beta;
}op[210005];
int n, q, m, i, rad, val1, val2, aux;
int main()
{
    cin >> q;
    while(q --)
    {
        cin >> n;
        m = 0;
        if(n == 3)
        {
            cout << "2" << "\n";
            cout << "3 2" << "\n";
            cout << "3 2" << "\n";
            continue;
        }
        if(n == 4)
        {
            cout << "3" << "\n";
            cout << "3 4" << "\n";
            cout << "4 2" << "\n";
            cout << "4 2" << "\n";
            continue;
        }
        val1 = n;
        rad = sqrt(n);
        if(rad * rad != n)val1 = rad + 1;
        else val1 = rad;
        rad = sqrt(val1);
        if(rad * rad != val1)val2 = rad + 1;
        else val2 = rad;
        if(n > 16)
        {
            for(i = 3; i <= n; i ++)
                if(i != val1 && i != val2 && i != n && i != 2)
                {
                    op[++ m].alfa = i;
                    op[m].beta = n;
                }
            op[++ m].alfa = n;
            op[m].beta = val1;
            op[++ m].alfa = n;
            op[m].beta = val1;
            op[++ m].alfa = val1;
            op[m].beta = val2;
            op[++ m].alfa = val1;
            op[m].beta = val2;
            aux = val2;
            while(aux > 1)
            {
                op[++ m].alfa = val2;
                op[m].beta = 2;
                if(aux % 2 == 0)aux = aux / 2;
                else aux = aux / 2 + 1;
            }
        }
        else
        {
            for(i = 3; i < n; i ++)
                {
                    op[++ m].alfa = i;
                    op[m].beta = n;
                }
            aux = n;
            while(aux > 1)
            {
                op[++ m].alfa = n;
                op[m].beta = 2;
                if(aux % 2 == 0)aux = aux / 2;
                else aux = aux / 2 + 1;
            }
        }
        cout << m << "\n";
        for(i = 1; i <= m; i ++)
            cout << op[i].alfa << " " << op[i].beta << "\n";
    }
    return 0;
}