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;
#define pb push_back
#define mp make_pair
#define fi first
#define se second


ll n;

void init() {
}

void input() {
    cin >> n;
}

ll sq (ll n) {
    ll l = 0, r = n + 1;

    while (l + 1 < r) {
        ll m = (l + r) / 2;

        if (m * m > n) {
            r = m;
        }

        else {
            l = m;
        }
    }

    if (l * l < n) {
        l++;
    }

    return l;
}

void solve() {
    int ans = 0;

    int inst = n;

    while (n > 2) {
        int newn = sq(n);

        ans += n - newn - 1;

        ans += 2;

        n = newn;
    }

    cout << ans << endl;

    n = inst;

    while (n > 2) {
        int newn = sq(n);

        for (int i = newn + 1; i < n; i++) {
            cout << i << ' ' << i + 1 << endl;
        }

        cout << n << ' ' << newn << endl;

        cout << n << ' ' << newn << endl;

        n = newn;
    }
}

void output() {
}

int main()
{
    ios_base::sync_with_stdio(false);

    int number_of_testcases = 1;
    cin >> number_of_testcases;
    while (number_of_testcases--) {
        init();

        input();

        solve();

        output();
    }

    return 0;
}