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:
#define _CRT_SECURE_NO_WARNINGS
#include <bits/stdc++.h>
#define all(x) (x).begin(), (x).end()
#define make_unique(x) sort(all((x))); (x).resize(unique(all((x))) - (x).begin())
#define D(x) cout << #x << " = " << x << endl
using namespace std;
using LL = long long;
using pii = pair<int, int>;

const int maxn = 3030;
constexpr int mod = 1000000007;
int n, t, nums[maxn];

inline int nxt() {
	int x;
	cin >> x;
	return x;
}

void solve() {
	n = nxt();
	vector<pii> ans;

	/*
	已经有 1 个 1 和 1 个 2， 生成 n - 2 个 1 即可，
	3...n 至多使用 n + 5 次操作。
	对于任意 x ，直接 x x + 1 则一次即可生成 1
	设在 x 处 次数为 log n
	1. >=x  (n - x) * 2
	2. < x  (x - 3) 
	log n + (n - x) * 2 + (x - 3) <= n + 5
	
	*/
	while (n > 2) {
		int s = (int)ceil(sqrt(n));
		for (int i = s + 1; i < n; ++i) {
			ans.emplace_back(i, i + 1);
		}
		ans.emplace_back(n, s);
		ans.emplace_back(n, s);
		n = s;
	}
	cout << ans.size() << "\n";
	for (auto x : ans) {
		cout << x.first << " " << x.second << "\n";
	}
}

int main() 
{
#ifdef DEBUG
	freopen("in.txt", "r", stdin);
#endif
	scanf("%d", &t);
	while (t--) {
		solve();
	}
	return 0;
}