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>
#define ri register int
#define ll long long
#define lson rt << 1, l, mid
#define rson rt << 1|1, mid+1, r
using namespace std;
void re(int &x){
    x = 0;
    int b = 0;
    char ch = getchar();
    while(ch < '0' || ch > '9'){
        if(ch == '-') b = 1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9'){
        x = (x << 1) + (x << 3) + ch - '0';
        ch = getchar();
    }
    if(b == 1)
        x *= -1;
}
const int maxn = 410000;
const int p = 998244353;
int n, t, q, k;
int a[maxn], b[maxn], cnt;
void out(){
    cout << cnt << endl;
    for(int i = 1;i <= cnt;i ++)
        cout << a[i] << " " << b[i] << endl;
}
void work(){
    for(int i = 3;i < n;i ++){
        if(i == 8) continue;
        cnt ++;
        a[cnt] = i;
        b[cnt] = n;
    }
    int x = n;
    while(x != 1){
        cnt ++;
        a[cnt] = n;
        b[cnt] = 8;
        x = ceil(x/8.0);
    }
    cnt ++;
    a[cnt] = 8;
    b[cnt] = 2;
    cnt ++;
    a[cnt] = 8;
    b[cnt] = 2;
    cnt ++;
    a[cnt] = 8;
    b[cnt] = 2;
    out();
}
int main(){
    re(t);
    while(t --){
        cnt = 0;
        re(n);
        if(n > 8) work();
        else{
            for(int i = 3;i < n;i ++){
                cnt ++;
                a[cnt] = i;
                b[cnt] = n;
            }
            int x = n;
            while(x != 1){
                cnt ++;
                a[cnt] = n;
                b[cnt] = 2;
                x = ceil(x/2.0);

            }
            out();
        }

    }

    return 0;
}