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]. 
#include<bits/stdc++.h>
using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const int N = 2e5 + 10;
vector<PII> ans, res;

void solve() {
    set<int> s; ans.clear(); res.clear();
    int n;
    scanf("%d", &n);
    int num = n;
    while(num != 1) {
        int a = floor(sqrt(1.0 * num));
        if(a == 1) a++;
        s.insert(a);
        ans.push_back({n, a});
        if(a != 2) ans.push_back({a, n});
        num = ceil(1.0 * num / a);
    }
    for(int i = 3; i < n; i++) {
        if(!s.count(i)) res.push_back({i, n});
    }
    printf("%d\n", res.size() + ans.size());
    for(int i = 0; i < res.size(); i++) {
        printf("%d %d\n", res[i].first, res[i].second);
    }
    for(int i = 0; i < ans.size(); i++) {
        printf("%d %d\n", ans[i].first, ans[i].second);
    }
}

int main() {
    // freopen("in.txt", "r", stdin);
    int t; cin >> t; while(t--)
    solve();
    return 0;
}