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 long double ld;
typedef pair<ll, ll> ii;
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define sqr(x) ((x) * (x))
#define X first
#define Y second
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) FOR (i, 0, n)
#define TRACE(x) cout << #x << " = " << x << endl
#define _ << " _ " <<
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define MAXN 200200

int n;
set<int> s;
vector<int> v;

void solve()
{
  s.clear();
  v.clear();
  scanf("%d", &n);
  for (int i = n; i > 2; i = ceil(sqrt(double(i)))) {
    s.insert(i);
    v.pb(i);
  }
  v.pb(2);
  s.insert(2);
  printf("%d\n", n - 1 - s.size() + 2 * (s.size() - 1));
  for (int i = 3; i <= n; i++) {
    if (!s.count(i)) {
      printf("%d %d\n", i, i + 1);
    }
  }
  for (int i = 0; i < v.size() - 1; i++) {
    printf("%d %d\n", v[i], v[i + 1]);
    printf("%d %d\n", v[i], v[i + 1]);
  }
}

int main()
{
  int t;
  scanf("%d", &t);
  while (t--) solve();
  return 0;
}
