Ceil Divisions

D. Ceil Divisions

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=lceil{frac{a_x}{a_y}} ceil)(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 ((1leq tleq 1000)) — the number of test cases.

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

It's guaranteed that the sum of n over test cases doesn't exceed 2⋅10⁵.

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 ((mleq n+5mleq n+5)) — the number of operations; next print m pairs of integers x and y ((1leq x,yleq 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=lceil{frac{a_3}{a_2}} ceil=2) and array a=[1,2,2];
2. choose 3, 2: (a_3=lceil{frac{a_3}{a_2}} ceil=1) and array a=[1,2,1].

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

一开始想的是取log2(n)个或者是log10(n)个不除,然后其余的除以n,全变成1 ,用n除以(n+1)/2等等,但是总是多几次

正解:

n用16(或15等)除若干次,变为1,16再用2除以4次,变成1

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MAXN=108,mod=1e9+7,inf=0x3f3f3f3f;
int main(){
    int t,n;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        if(n<=16){
            printf("%d
",n-3+(int)ceil(log2(n)));
            for(int i=3;i<n;++i)printf("%d %d
",i,n);
            for(int i=n;i>1;i=(i+1)/2)
                printf("%d 2
",n);
        }else{
            printf("%d
",n-4+4+(int)ceil(log2(n)/4.0));
            for(int i=3;i<16;++i)printf("%d %d
",i,n);
            for(int i=17;i<n;++i)printf("%d %d
",i,n);
            for(int i=n;i>1;i=(i+15)/16)printf("%d 16
",n);
            for(int i=0;i<4;++i)printf("16 2
");
        }
    }
    return 0;
}