#include <bits/stdc++.h>

#define fast ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define SIZE 500000
using namespace std;
typedef long long ll;
const ll INF = 1e10;
ll t = 1, n, m, k;
ll temp;
string s;
ll arr[SIZE];


int main() {
    fast;
    cin >> t;
    while (t--) {
        cin >> n;
        vector<ll> special;
        ll curr = n;
        while (curr > 2) {
            special.push_back(ceil(curr));
            curr = ceil(sqrt(curr));
        }
        vector<pair<ll, ll>> res;
        for (int i = 3; i <= n - 1; ++i) {
            bool found = false;
            for (auto j : special) if (j == i) found = true;
            if (!found) res.push_back({i, n});
        }
        sort(special.begin(), special.end());
        for (int i = special.size() - 1; i >= 1; --i) {
            res.push_back({special[i], special[i - 1]});
            res.push_back({special[i], special[i - 1]});
        }
        res.push_back({special[0], 2});
        res.push_back({special[0], 2});
        cout << res.size() << "\n";
        for (auto p : res) cout << p.first << " " << p.second << "\n";
    }
}


### Prompt

Construct a CPP code solution to the problem outlined:

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]. 

### Response

```cpp
#include <bits/stdc++.h>

#define fast ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define SIZE 500000
using namespace std;
typedef long long ll;
const ll INF = 1e10;
ll t = 1, n, m, k;
ll temp;
string s;
ll arr[SIZE];


int main() {
    fast;
    cin >> t;
    while (t--) {
        cin >> n;
        vector<ll> special;
        ll curr = n;
        while (curr > 2) {
            special.push_back(ceil(curr));
            curr = ceil(sqrt(curr));
        }
        vector<pair<ll, ll>> res;
        for (int i = 3; i <= n - 1; ++i) {
            bool found = false;
            for (auto j : special) if (j == i) found = true;
            if (!found) res.push_back({i, n});
        }
        sort(special.begin(), special.end());
        for (int i = special.size() - 1; i >= 1; --i) {
            res.push_back({special[i], special[i - 1]});
            res.push_back({special[i], special[i - 1]});
        }
        res.push_back({special[0], 2});
        res.push_back({special[0], 2});
        cout << res.size() << "\n";
        for (auto p : res) cout << p.first << " " << p.second << "\n";
    }
}


```