Problem: There are n cats in a line, labeled from 1 to n, with the i-th cat at position i. They are bored of gyrating in the same spot all day, so they want to reorder themselves such that no cat is in the same place as before. They are also lazy, so they want to minimize the total distance they move. Help them decide what cat should be at each location after the reordering.

For example, if there are 3 cats, this is a valid reordering: [3, 1, 2]. No cat is in its original position. The total distance the cats move is 1 + 1 + 2 = 4 as cat 1 moves one place to the right, cat 2 moves one place to the right, and cat 3 moves two places to the left.

Input

The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.

The first and only line of each test case contains one integer n (2 ≤ n ≤ 100) — the number of cats.

It can be proven that under the constraints of the problem, an answer always exist. 

Output

Output t answers, one for each test case. Each answer consists of n integers — a permutation with the minimum total distance. If there are multiple answers, print any.

Example

Input


2
2
3


Output


2 1 
3 1 2 

Note

For the first test case, there is only one possible permutation that satisfies the conditions: [2, 1].

The second test case was described in the statement. Another possible answer is [2, 3, 1].

Solution (in CPP):

Solution:
#include <iostream>
#include <bits/stdc++.h>
using namespace std;

#define MAX     1000000
#define pb       push_back
#define ppb      pop_back
#define FOR(i,s,e)      for(int i=s; i<=e; i++) 
#define RFOR(i,e,s)     for(int i=e; i>=s; i--)
#define lli     long long int
#define mod       1000000007
#define ppc     __builtin_popcount
#define ppcll     __builtin_popcountll
#define gs      getline(cin, s);
#define c_yes   std::cout << "YES" << std::endl;
#define c_no   std::cout << "NO" << std::endl;
#define ALL(x)      x.begin() , x.end()
typedef   pair<int, int> pi;
typedef   std::vector<int> vi;
typedef   std::vector<vector<int>> vvi;
typedef   std::vector<lli> vlli;
typedef   std::vector<char> vc;
typedef   std::vector<string> vs;
typedef   std::vector<pi> vpi;
typedef   priority_queue<int> max_heap_int;
typedef   priority_queue<string> max_heap_string;
typedef   priority_queue<int, vi, greater<int>> min_heap_int;


// class Timer { private: chrono::time_point <chrono::steady_clock> Begin, End; public: Timer () : Begin(), End ()
// { Begin = chrono::steady_clock::now(); } ~Timer () { End = chrono::steady_clock::now();
// cout << "\nDuration: " << ((chrono::duration <double>)(End - Begin)).count() << "s\n"; } } t;

template <class T>
void countSort(vector<T>&arr){
    
    T max = *max_element(arr.begin(), arr.end());
    T min = *min_element(arr.begin(), arr.end());
    int range = max - min + 1;
 
    vector<int> count(range);
    std::vector<T> output(arr.size());
    for (int i = 0; i < arr.size(); i++)
        count[arr[i] - min]++;
 
    for (int i = 1; i < count.size(); i++)
        count[i] += count[i - 1];
 
    for (int i = arr.size() - 1; i >= 0; i--) {
        output[count[arr[i] - min] - 1] = arr[i];
        count[arr[i] - min]--;
    }
 
    for (int i = 0; i < arr.size(); i++)
        arr[i] = output[i];
    
   
}

template <class T>
bool isPal(T n){
    string s = n;
    int i=0, j=s.length()-1;
    while(i<=j){
        if(s[i] != s[j])
            return false;
        i++;j--;
    }
    return true;
}

bool isPrime(int n){
    if(n<=1 || n==4) return false;
    if(n<=3) return true;
    
    for(int i=2; i*i<=n; i++){
        if(n%i==0)return false;
    }
    return true;
}

lli power(lli a, lli p) {
    if(p == 0) return 1;
    
    if(p % 2) 
        return (a % mod * power(a , p-1) % mod) % mod;
    
    return power((a % mod * a % mod) % mod , p/2) % mod;
}
lli modsum(lli a , lli b) {return (a % mod + b % mod + mod) % mod;}
int tc = 0;

const int dx4[4] = {1,-1,0,0} , dy4[4] = {0,0,1,-1};

const lli N = 1e9+1;
vector<bool> squareFree(N , true);

bool isSquare(lli n) {
    lli s = (lli) sqrt(n);
    return s * s == n;
}

bool square_free(lli n) {

    if(n == 1) return false;
    
    for(lli i=2; i*i <= n; i++) 
    {
        if(n % i == 0 and (isSquare(i) || isSquare(n/i)))  return false;
    }
    return true;
}


void solve(){  
    
    int n;
    cin >> n;
    
    vi a(n+1);
    FOR(i,1,n) a[i] = i;
    
    if(n % 2 == 0) {
        for(int i=1; i<n; i+=2) 
            swap(a[i] , a[i+1]);
    }
    
    else {
        for(int i=1; i<n-2; i+=2) 
            swap(a[i] , a[i+1]);
        
        swap(a[n] , a[n-2]);
        swap(a[n] , a[n-1]);
    }
    
    FOR(i,1,n) cout << a[i] << " ";
    cout << "\n";
    
} 

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    
    
    
    int t;
    cin >> t;
    
    while(t--)
    solve();
	
	return 0;
}