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;
#define rep(i, n) for (int i = 0; (i) < (int) (n); (i) ++)
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()
#define dump(x) cerr << #x << " = " << x << '\n';
using ll = long long;
template<typename T, typename U> ostream& operator << (ostream& os, const pair<T, U>& p) { os << p.first << " " << p.second; return os; }
template<typename T> ostream& operator << (ostream& os, const vector<T>& vec) { for (int i = 0; i < vec.size(); i ++) { os << vec[i] << (i + 1 == vec.size() ? "" : " "); } return os; }
template<typename T> bool chmin(T &a, const T& b) { if (a > b) { a = b; return true; } return false; }
template<typename T> bool chmax(T &a, const T& b) { if (a < b) { a = b; return true; } return false; }

void solve() {
        int n;
        cin >> n;
        if (n <= 4) {
                if (n == 3) {
                        cout << "2\n3 2\n3 2\n";
                        return;
                } else if (n == 4) {
                        cout << "3\n3 4\n4 2\n4 2\n";
                        return;
                }
        }
        if (n <= 60) {
                vector<pair<int, int>> ans;
                for (int i = 3; i <= n - 1; i ++) {
                        if (i != 4) {
                                ans.emplace_back(i, n);
                        }
                }
                int val = n;
                while (val > 1) {
                        ans.emplace_back(n, 4);
                        val = (val + 3) / 4;
                }
                ans.emplace_back(4, 2);
                ans.emplace_back(4, 2);
                cout << ans.size() << '\n';
                assert(ans.size() <= n + 5);
                rep(i, ans.size()) {
                        cout << ans[i] << '\n';
                }
                return;
        }
        vector<pair<int, int>> ans;
        for (int i = 3; i <= n - 1; i ++) {
                if (i != 4 && i != 60) {
                        ans.emplace_back(i, n);
                }
        }
        int val = n;
        while (val > 1) {
                ans.emplace_back(n, 60);
                val = (val + 59) / 60;
        }
        val = 60;
        while (val > 1) {
                ans.emplace_back(60, 4);
                val = (val + 3) / 4;
        }
        ans.emplace_back(4, 2);
        ans.emplace_back(4, 2);
        assert(ans.size() <= n + 5);
        cout << ans.size() << '\n';
        rep(i, ans.size()) {
                cout << ans[i] << '\n';
        }
}

int main() {
        cin.tie(nullptr);
        ios::sync_with_stdio(false);
        int t; cin >> t;
        // int t = 1;
        while (t --) {
                solve();
        }
        return 0;
}