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]. 
//
//  main.cpp
//  edu codeforces 101 A
//
//  Created by Joshua Koo on 12/28/20.
//
 
#include <stdio.h>
#include <algorithm>
#include <limits.h>
#include <map>
#include <set>
#include <math.h>
#include <stdlib.h>
#include <vector>
#include <set>
#include <string.h>
FILE *in = stdin;
FILE *out = stdout;
using namespace std;
int t;
int main() {
    fscanf(in, "%d", &t);
    for (int cases = 1; cases <= t; cases++) {
        int n;
        fscanf(in, "%d", &n);
        vector<double> v;
        map<double, int> m;
        int ans;
        int cnt = 0;
        double current = (double)n;
        while (current >= 2) {
            v.push_back(current);
            cnt++;
            m[current] = 1;
            if (current == 2) break;
            current = ceil(sqrt(current));
        }
        m[1] = 1;
        v.push_back(1);
        cnt++;
        ans = n - cnt;
        for (int i = 2; i <= v.size() - 1; i++) {
            ans += 2;
        }
        fprintf(out, "%d\n", ans);
        for (double i = 1; i <= n; i++) {
            if (m[i] == 0) {
                fprintf(out, "%d %d\n", (int)i, (int)i + 1);
            }
        }
        for (int i = 0; i < v.size() - 2; i++) {
            fprintf(out, "%d %d\n%d %d\n", (int)v[i], (int)v[i+1], (int)v[i], (int)v[i+1]);
        }
    }
    return 0;
}
