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:
#ifdef DBG_MACRO_NO_WARNING
#include <dbg.h>
#else
#define dbg(...) ((void)0)
#endif
#include <bits/stdc++.h>
#define all(c) begin(c), end(c)
#define isz(c) (int)(c).size()

using namespace std;
using ll = long long;
using ii = pair<int, int>;
using vi = vector<int>;

inline int ceil(int a, int b) { return (a + b - 1) / b; }

int bs(int l, int r, function<bool(int)> p)
{
    // 0 0 0 0 1 1
    while (l < r)
    {
        int m = l + (r - l) / 2;
        if (p(m))
            r = m;
        else
            l = m + 1;
    }
    return l;
}

void solve(int n, vector<ii> &ans)
{
    if (n == 2)
        return;
    int i = bs(1, n - 1, [n](int x) { return x >= ceil(n, x); });

    for (int j = i + 1; j < n; ++j)
        ans.emplace_back(j, n);
    ans.emplace_back(n, i);
    ans.emplace_back(n, i);
    solve(i, ans);
}

int main(void)
{
    ios::sync_with_stdio(false), cin.tie(NULL);
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        vector<ii> ans;
        solve(n, ans);

        cout << ans.size() << endl;
        for (auto [x, y] : ans)
            cout << x << " " << y << endl;
        cout << endl;
    }
    return 0;
}