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;
using ull = unsigned long long;
constexpr ll MOD = 1000 * 1000 * 1000 + 7;
constexpr int ALPHABET_SIZE = 26;
constexpr ll INF = 1e18;

int main() {
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    ios_base::sync_with_stdio(false);

    int t;
    cin >> t;
    for (int it = 0; it < t; ++it) {
        int n;
        cin >> n;
        vector<bool> used(n, true);
        used[0] = used[1] = false;
        int k = n;
        vector<int> sqrts;
        while (k > 2) {
            used[k - 1] = false;
            sqrts.push_back(k);
            int sq = ceil(sqrt(k + 0.));
            if (sq * sq < k) {
                ++sq;
            }
            k = sq;
        }
        vector<pair<int, int>> ans;
        for (int i = 0; i < n; ++i) {
            if (used[i]) {
                ans.emplace_back(i + 1, n);
            }
        }
        for (size_t i = 0; i < sqrts.size(); ++i) {
            int cur = sqrts[i], nxt = (i + 1 < sqrts.size()) ? sqrts[i + 1] : 2;
            ans.emplace_back(cur, nxt);
            ans.emplace_back(cur, nxt);
        }
        cout << ans.size() << "\n";
        for (auto [i, j] : ans) {
            cout << i << " " << j << "\n";
        }
    }

    return 0;
}