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]. 
#include<bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#define ll long long
#define pb push_back
#define ull unsigned long long
#define db double
#define lowbit(x) x&-x 
#define mem(x) memset(x,0,sizeof x)
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define vi vector<int>
#define pii pair<int,int>
const ll mod = 1000000007;
int pow_mod(int a,int b){ int ret=1;
for(;b;b>>=1,a=1LL*a*a%mod)if(b&1)ret=1LL*ret*a%mod;
return ret; }//快速幂
int gcd(int a, int b){    return a%b==0? b: gcd(b, a%b); }//递归求最大公约数
inline ll read() {ll x = 0; int t=1;char ch = getchar();
while(ch<'0'||ch>'9'){ if(ch=='-') t=-1; ch = getchar();}
while(ch>='0'&&ch<='9'){ x=x*10+ch-'0'; ch =getchar(); }
return x*t; } //快读

const int N = 100100;


int a[N];
int b[N];
void solve(){
    int n;
    cin >> n;
    vector<pii>p;
    int k = n;
    while(k > 2){
        int t = sqrt(k);
        if(t * t != k) t ++;
        for(int i = k - 1;i > t;i --){
            p.push_back({i,k});
        }
        p.push_back({k,t});
        p.push_back({k,t});
        k = t;
    }

    cout << p.size() << endl;
    for(auto it : p){
        cout << it.first << ' ' << it.second << endl;
    }
}

int main()
{
    // freopen("B.in","r",stdin);
    // freopen("B.out","w",stdout);
    int t;
    cin >> t;
    while(t --)
	solve();
	return 0;
}