C. Candies

Problem

This problem is about candy. Initially, you only have $1$ candy, and you want to have exactly $n$ candies.

You can use the two following spells in any order at most $40$ times in total.

• Assume you have $x$ candies now. If you use the first spell, then $x$ candies become $2x-1$ candies.
• Assume you have $x$ candies now. If you use the second spell, then $x$ candies become $2x+1$ candies.

Construct a sequence of spells, such that after using them in order, you will have exactly $n$ candies, or determine it’s impossible.

Input

Each test contains multiple test cases. The first line contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases. Their description follows.

Each test case contains one line with a single integer $n$ ($2\le n\le 10^9$) — the required final number of candies.

Output

For each test case, output the following.

If it’s possible to eventually have $n$ candies within $40$ spells, in the first line print an integer $m$ ($1\le m\le 40$), representing the total number of spells you use.

In the second print $m$ integers $a_1,a_2,\dots ,a_m$ ($a_i$ is $1$ or $2$) separated by spaces, where $a_i=1$ means that you use the first spell in the $i$-th step, while $a_i=2$ means that you use the second spell in the $i$-th step.

Note that you do not have to minimize $m$, and if there are multiple solutions, you may output any one of them.

If it’s impossible, output $-1$ in one line.

Example

Input

4
2
3
7
17

Output

-1
1
2 
2
2 2 
4
2 1 1 1 

Note

For $n=3$, you can just use the second spell once, and then have $2\cdot 1+1=3$ candies.

For $n=7$, you can use the second spell twice. After the first step, you will have $3$ candies. And after the second step, you will have $2\cdot 3+1=7$ candies.

Code

// #include <iostream>
// #include <algorithm>
// #include <cstring>
// #include <stack>//栈
// #include <deque>//堆/优先队列
// #include <queue>//队列
// #include <map>//映射
// #include <unordered_map>//哈希表
// #include <vector>//容器，存数组的数，表数组的长度
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

void solve()
{
    ll n;
    cin>>n;

    vector<ll> a;
    if(n%2==0)
    {
        cout<<-1<<endl;
        return;
    }

    while(n>1)
    {
        if(((n-1)/2)&1)
        {
            a.push_back(2);
            n=(n-1)/2;
        }
        else if(((n+1)/2)&1)
        {
            a.push_back(1);
            n=(n+1)/2;
        }
    }

    cout<<a.size()<<endl;
    reverse(a.begin(),a.end());
    for(auto &t:a)
        cout<<t<<" ";
    cout<<endl;
}

int main()
{
    ll t;
    cin>>t;
    while(t--) solve();
    
    return 0;
}