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]. 
import sys
input = lambda: sys.stdin.readline().rstrip("\r\n")
inp = lambda: list(map(int,sys.stdin.readline().rstrip("\r\n").split()))
#______________________________________________________________________________________________________
from math import *
# from bisect import *
# from heapq import *
# from collections import defaultdict as dd
# from collections import OrderedDict as odict
# from collections import Counter as cc
# from collections import deque
# sys.setrecursionlimit(2*(10**5)+100) this is must for dfs
# mod = 10**9+7; md = 998244353
# ______________________________________________________________________________________________________
# segment tree for range minimum query
# sys.setrecursionlimit(10**5)
# n = int(input())
# a = list(map(int,input().split()))
# st = [float('inf') for i in range(4*len(a))]
# def build(a,ind,start,end):
#   if start == end:
#       st[ind] = a[start]
#   else:
#       mid = (start+end)//2
#       build(a,2*ind+1,start,mid)
#       build(a,2*ind+2,mid+1,end)
#       st[ind] = min(st[2*ind+1],st[2*ind+2])
# build(a,0,0,n-1)
# def query(ind,l,r,start,end):
#   if start>r or end<l:
#       return float('inf')
#   if l<=start<=end<=r:
#       return st[ind]
#   mid = (start+end)//2
#   return min(query(2*ind+1,l,r,start,mid),query(2*ind+2,l,r,mid+1,end))
# ______________________________________________________________________________________________________
# Checking prime in O(root(N))
# def isprime(n):
#     if (n % 2 == 0 and n > 2) or n == 1: return 0
#     else:
#         s = int(n**(0.5)) + 1
#         for i in range(3, s, 2):
#             if n % i == 0:
#                 return 0
#         return 1
# def lcm(a,b):
#   return (a*b)//gcd(a,b)
# ______________________________________________________________________________________________________
# nCr under mod
# def C(n,r,mod):
#   if r>n:
#       return 0
#   num = den = 1
#   for i in range(r):
#       num = (num*(n-i))%mod
#       den = (den*(i+1))%mod
#   return (num*pow(den,mod-2,mod))%mod
# M = 10**5 +10
# ______________________________________________________________________________________________________
# For smallest prime factor of a number
# M = 1000010
# pfc = [i for i in range(M)]
# def pfcs(M):
#   for i in range(2,M):
#       if pfc[i]==i:
#           for j in range(i+i,M,i):
#               if pfc[j]==j:
#                   pfc[j] = i
#   return
# pfcs(M)
# ______________________________________________________________________________________________________
tc = 1
tc, = inp()
for _ in range(tc):
    n, = inp()
    m = n
    a = [n]
    while(m!=2):
        t = ceil(m**0.5)
        a.append(t)
        m = t
    s = set(a)
    ans = [i for i in range(1,n+1)]
    res = []
    for j in range(3,n):
        if j not in s:
            res.append((j,j+1))
    for i in range(1,len(a)):
        res.append((a[i-1],a[i]))
        res.append((a[i-1],a[i]))
    print(len(res))
    for i,j in res:
        print(i,j)