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;
 
using ll = long long;
using str = string; 
 
using pi = pair<int,int>;
using pl = pair<ll,ll>;
using vi = vector<int>;
using vl = vector<ll>;
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>; 
 
#define sz(x) (int)(x).size()
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend() 
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define lb lower_bound 
#define ub upper_bound 
 
constexpr int setbits(int x) { return __builtin_popcount(x); } 
constexpr int bits(int x) { return 32-__builtin_clz(x); } 
str binary(int x) { return bitset<32>(x).to_string(); } 
int decimal(str x) { return stoi(x,nullptr,2); } 
ll cdiv(ll a, ll b) { return a/b+((a^b)>0&&a%b); } 
ll fdiv(ll a, ll b) { return a/b-((a^b)<0&&a%b); } 
 
const int nax = INT_MAX;
const int nim = INT_MIN;	
const int mod = 1000000007;

void solve(){
	int n; cin >> n;
	vector<pair<int, int> > ans;
	vector<int> v(n + 1);
	for(int i = 1; i <= n; i++) {
		v[i] = i;
	}
	for(int i = n - 1; i >= 3; i--) {
		if((v[n] + v[i] - 1) / v[i] >= v[i]) {
			v[n] = (v[n] + v[i] - 1) / v[i];
			ans.push_back({n, i});
		}
		ans.push_back({i, n});
	}
	while(v[n] > 1) {
		ans.push_back({n, 2});
		v[n] = (v[n] + 1) / 2;
	}
	assert(ans.size() <= n + 5);
	cout << ans.size() << endl;
	for(auto val: ans) {
		cout << val.first << " " << val.second << endl;
	}
}

int main(){
	ios_base::sync_with_stdio(false), cin.tie(nullptr);
	int t = 1;
	cin >> t;
	while(t--)
		solve();
}