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 JAVA):

Solution:
import java.util.*;
import java.io.*;

public class CFA {
  BufferedReader br;
  PrintWriter out;
  StringTokenizer st;
  boolean eof;
  private static final long MOD = 1000L * 1000L * 1000L + 7;
  private static final int[] dx = {0, -1, 0, 1};
  private static final int[] dy = {1, 0, -1, 0};
  private static final String yes = "Yes";
  private static final String no = "No";

  void solve() {
    int T = nextInt();
//    int T = 1;
    for (int i = 0; i < T; i++) {
      helper();
    }
  }

  List<Integer> left;
  List<Integer> right;
  int[] arr;
  void helper() {
    int n = nextInt();
    left = new ArrayList<>();
    right = new ArrayList<>();
    arr = new int[n];
    for (int i = 0; i < n; i++) {
      arr[i] = i + 1;
    }

    Map<Integer, Integer> hm = new HashMap<>();
    int start = 2;
    for (int i = 0; i < 4; i++) {
      hm.put(start * start, start);
      if (start * start >= n) {
        break;
      }
      start = start * start;
    }

    if (!hm.containsKey(n)) {
      hm.put(n, start);
    }

    for (int i = 3; i < n; i++) {
      if (!hm.containsKey(i)) {
        add(i - 1, i);
      }
    }

    for (int i = n; i >= 3; i--) {
      if (hm.containsKey(i)) {
        while (arr[i - 1] > 1) {
          add(i - 1, hm.get(i) - 1);
        }
      }
    }

    outln(left.size());
    for (int i = 0; i < left.size(); i++) {
      outln(left.get(i) + " " + right.get(i));
    }
  }

  void add(int l, int r) {
    arr[l] = (arr[l] + arr[r] - 1) / arr[r];
    left.add(l + 1);
    right.add(r + 1);
  }

  void shuffle(long[] a) {
    int n = a.length;
    for(int i = 0; i < n; i++) {
      int r = i + (int) (Math.random() * (n - i));
      long tmp = a[i];
      a[i] = a[r];
      a[r] = tmp;
    }
  }

  long gcd(long a, long b) {
    while(a != 0 && b != 0) {
      long c = b;
      b = a % b;
      a = c;
    }
    return a + b;
  }
  private void outln(Object o) {
    out.println(o);
  }
  private void out(Object o) {
    out.print(o);
  }
  private void formatPrint(double val) {
    outln(String.format("%.9f%n", val));
  }
  public CFA() {
    br = new BufferedReader(new InputStreamReader(System.in));
    out = new PrintWriter(System.out);
    solve();
    out.close();
  }
  public static void main(String[] args) {
    new CFA();
  }

  public long[] nextLongArr(int n) {
    long[] res = new long[n];
    for(int i = 0; i < n; i++)
      res[i] = nextLong();
    return res;
  }
  public int[] nextIntArr(int n) {
    int[] res = new int[n];
    for(int i = 0; i < n; i++)
      res[i] = nextInt();
    return res;
  }
  public String nextToken() {
    while (st == null || !st.hasMoreTokens()) {
      try {
        st = new StringTokenizer(br.readLine());
      } catch (Exception e) {
        eof = true;
        return null;
      }
    }
    return st.nextToken();
  }
  public String nextString() {
    try {
      return br.readLine();
    } catch (IOException e) {
      eof = true;
      return null;
    }
  }
  public int nextInt() {
    return Integer.parseInt(nextToken());
  }
  public long nextLong() {
    return Long.parseLong(nextToken());
  }
  public double nextDouble() {
    return Double.parseDouble(nextToken());
  }
}