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;
#define mem(x) memset(x, 0, sizeof x)
#define all(x) (x).begin(), (x).end()
#define endl "\n"
typedef long long ll;
inline ll rint() { ll x = 0, f = 1; char c = getchar(); while (c<'0' || c>'9') { if (c == '-')f = -1; c = getchar(); }while (c >= '0'&&c <= '9') { x = x * 10 + c - '0'; c = getchar(); }return x * f; }
inline ll gcd(ll a, ll b) { return (b) ? gcd(b, a%b) : a; }
inline ll lcm(ll a, ll b) { return a / gcd(a, b)*b; }
inline ll pow(ll a, ll b, ll q) { ll rtn = 1; while (b) { if (b & 1) rtn = rtn * a%q; a = a * a%q; b >>= 1; } return rtn; }
inline ll mysqrt(ll n) { ll x = sqrt(n); while (x*x < n) x++; while (x*x > n) x--; return x; }

void solve() {
  ll n=rint();
  vector<pair<ll,ll>> v;
  vector<ll> lishu = {2,3,5,22,448};
  for(int i=4;i<=n-1;i++) if(find(all(lishu),i)==lishu.end()) v.emplace_back(i,n);
  while(*lishu.rbegin()>=n) lishu.pop_back();
  lishu.push_back(n);
  for(int i=lishu.size()-1;i>=1;i--) v.emplace_back(lishu[i],lishu[i-1]),v.emplace_back(lishu[i],lishu[i-1]);
  cout<<v.size()<<endl;
  for(auto i:v) cout<<i.first<<" "<<i.second<<endl;
}

int main() {
  ll t=rint(); while(t--) solve();
//  solve();
  return 0;
}