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 sz(x) (int)((x).size())
# define f first
# define s second
# define pb push_back
# define all(x) x.begin(), x.end()

using namespace std;

using ll = long long;
using pii = pair<int, int>;

const int mod = 1e9 + 7, N = 1e6 + 5;

void solve() {
    int n;
    cin >> n;
    /*
        n - 4 + 5 + 4


        1 2 [3 4 5 6]
    */

    vector<pii> ans;
    for (int i = 3; i < n; ++i) {
        if (i == 15 && n > 15) {
            continue;
        }
        ans.push_back(make_pair(i, n));
    }

    int z = n;
    while (z > 1) {
        if (n > 15) {
            z = (z + 14) / 15;
            ans.push_back(make_pair(n, 15));
        } else {
            z = (z + 1) / 2;
            ans.push_back(make_pair(n, 2));
        }
    }

    if (n > 15) {
        ans.push_back(make_pair(15, 2));
        ans.push_back(make_pair(15, 2));
        ans.push_back(make_pair(15, 2));
        ans.push_back(make_pair(15, 2));
    }

    cout << sz(ans) << "\n";
    for (auto i : ans) {
        cout << i.first << " " << i.second << "\n";
    }


}

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