Problem: Gildong is developing a game consisting of n stages numbered from 1 to n. The player starts the game from the 1-st stage and should beat the stages in increasing order of the stage number. The player wins the game after beating the n-th stage.

There is at most one checkpoint on each stage, and there is always a checkpoint on the 1-st stage. At the beginning of the game, only the checkpoint on the 1-st stage is activated, and all other checkpoints are deactivated. When the player gets to the i-th stage that has a checkpoint, that checkpoint is activated.

For each try of a stage, the player can either beat the stage or fail the stage. If they beat the i-th stage, the player is moved to the i+1-st stage. If they fail the i-th stage, the player is moved to the most recent checkpoint they activated, and they have to beat the stages after that checkpoint again.

For example, assume that n = 4 and the checkpoints are on the 1-st and 3-rd stages. The player starts at the 1-st stage. If they fail on the 1-st stage, they need to retry the 1-st stage because the checkpoint on the 1-st stage is the most recent checkpoint they activated. If the player beats the 1-st stage, they're moved to the 2-nd stage. If they fail it, they're sent back to the 1-st stage again. If they beat both the 1-st stage and the 2-nd stage, they get to the 3-rd stage and the checkpoint on the 3-rd stage is activated. Now whenever they fail on the 3-rd stage, or the 4-th stage after beating the 3-rd stage, they're sent back to the 3-rd stage. If they beat both the 3-rd stage and the 4-th stage, they win the game.

Gildong is going to build the stages to have equal difficulty. He wants you to find any series of stages and checkpoints using at most 2000 stages, where the [expected number](https://en.wikipedia.org/wiki/Expected_value) of tries over all stages is exactly k, for a player whose probability of beating each stage is exactly \cfrac{1}{2}.

Input

Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 50).

Each test case contains exactly one line. The line consists of a single integer k (1 ≤ k ≤ 10^{18}) — the expected number of tries over all stages Gildong wants to set for a player whose probability of beating each stage is exactly \cfrac{1}{2}.

Output

For each test case, print -1 if it's impossible to construct such a series of stages and checkpoints using at most 2000 stages.

Otherwise, print two lines. The first line should contain a single integer n (1 ≤ n ≤ 2000) – the number of stages. The second line should contain n integers, where the i-th integer represents whether the i-th stage has a checkpoint. The i-th integer should be 0 if the i-th stage doesn't have a checkpoint, and 1 if it has a checkpoint. Note that the first integer must be 1 according to the description.

Example

Input


4
1
2
8
12


Output


-1
1
1
4
1 1 1 1
5
1 1 0 1 1

Note

In the first and the second case, we can see that the 'easiest' series of stages is to have 1 stage with a checkpoint. This already requires 2 tries in expectation, so it is impossible to make it to require only 1 try.

In the third case, it takes 2 tries in expectation to beat each stage, and the player can always retry that stage without falling back to one of the previous stages if they fail it. Therefore the total expected number of tries is 8. Note that there exists an answer with fewer stages, but you are not required to minimize the number of stages used. 

Solution (in JAVA):

Solution:
import java.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.math.RoundingMode;
import java.util.*;
import java.util.concurrent.LinkedBlockingDeque;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.LinkedList;


// graph, dfs,bfs, get connected components,iscycle, isbipartite, dfs on trees



public class scratch_25 {
    static class Graph{

        public static class Vertex{
            HashMap<Integer,Integer> nb= new HashMap<>();     // for neighbours of each vertex
        }

        public static HashMap<Integer,Vertex> vt;   // for vertices(all)

        public Graph(){
            vt= new HashMap<>();
        }

        public static int numVer(){
            return vt.size();
        }

        public static boolean contVer(int ver){
            return vt.containsKey(ver);
        }

        public static void addVer(int ver){
            Vertex v= new Vertex();
            vt.put(ver,v);
        }






        public static void addEdge(int ver1, int ver2, int weight){
            if(!vt.containsKey(ver1) || !vt.containsKey(ver2)){
                return;
            }
            Vertex v1= vt.get(ver1);
            Vertex v2= vt.get(ver2);
            v1.nb.put(ver2,weight);                                 // if previously there is an edge, then this replaces that edge
            v2.nb.put(ver1,weight);
        }

        public static void delEdge(int ver1, int ver2){

            if(!vt.containsKey(ver1) || !vt.containsKey(ver2)){
                return;
            }
            vt.get(ver1).nb.remove(ver2);
            vt.get(ver2).nb.remove(ver1);
        }
        public static void delVer(int ver){
            if(!vt.containsKey(ver)){
                return;
            }
            Vertex v1= vt.get(ver);
            ArrayList<Integer> arr= new ArrayList<>(v1.nb.keySet());
            for (int i = 0; i <arr.size() ; i++) {
                int s= arr.get(i);
                vt.get(s).nb.remove(ver);
            }
            vt.remove(ver);
        }
           static boolean done[];
            static int parent[];
            static ArrayList<Integer>vals= new ArrayList<>();


        public static boolean isCycle(int i){
            Stack<Integer>stk= new Stack<>();

            stk.push(i);
            while(!stk.isEmpty()){
                int x= stk.pop();
                vals.add(x);
             //   System.out.print("current="+x+" stackinit="+stk);
                if(!done[x]){
                    done[x]=true;
                }
                else if(done[x] ){
                    return true;
                }

                ArrayList<Integer>ar= new ArrayList<>(vt.get(x).nb.keySet());
                for (int j = 0; j <ar.size() ; j++) {
                   if(parent[x]!=ar.get(j)){
                       parent[ar.get(j)]=x;
                    stk.push(ar.get(j));
                   }
                }
               // System.out.println(" stackfin="+stk);
            }
            return false;
        }

    static HashMap<Integer, Set<Integer>>map;
        static int color[];
        public static void bfs(int v){
            Queue<Integer>q= new LinkedList<>();
            q.add(v);
            while(!q.isEmpty()){
                int x= q.poll();
                done[x]=true;
                ArrayList<Integer>ar= new ArrayList<>(vt.get(x).nb.keySet());
                Set<Integer>set= new HashSet<>();
                if(!map.containsKey(color[x])){
                    set= new HashSet<>();
                    map.put(color[x],set);
                }
                else{
                    set= map.get(color[x]);
                }
                for (int i = 0; i <ar.size() ; i++) {
                    int d= ar.get(i);
                    if(color[d]!=color[x]){
                    set.add(color[d]);}
                    if(!map.containsKey(color[d])){
                        Set<Integer>s= new HashSet<>();
                        s.add(color[x]);
                        map.put(color[d],s);
                    }
                    else{
                        Set<Integer>s= map.get(color[d]);
                        if(color[d]!=color[x]){
                        s.add(color[x]);
                        map.put(color[d],s);
                    }}
                    if(!done[d]){
                        q.add(d);
                    }
                }
                 map.put(color[x],set);
            }
        }
        }
        



    // int count=0;
    //static long count=0;
    static class Reader {
        static BufferedReader reader;
        static StringTokenizer tokenizer;

        /**
         * call this method to initialize reader for InputStream
         */
        static void init(InputStream input) {
            reader = new BufferedReader(
                    new InputStreamReader(input));
            tokenizer = new StringTokenizer("");
        }

        /**
         * get next word
         */
        static String next() throws IOException {
            while (!tokenizer.hasMoreTokens()) {
                //TODO add check for eof if necessary
                tokenizer = new StringTokenizer(
                        reader.readLine());
            }
            return tokenizer.nextToken();
        }

        static int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        static double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        static long nextLong() throws IOException {
            return Long.parseLong(next());
        }
    }
    static class Pair implements Comparable<Pair>{
        long t;
        long left;
        long right;
        public Pair(long t,long left, long right){
            this.t=t;
            this.left=left;
            this.right= right;
        }
        @Override
        public int compareTo(Pair o){
            return (int)this.left-(int)o.left;
            
        }
        @Override
        public boolean equals(Object me) {
//            Pair binMe = (Pair)me;
//            if(this.x==binMe.x && this.y==binMe.y)
//                return true;
//            else
//                return false;
            return false;
        }

        @Override
        public int hashCode() {
            return (int)this.left + (int)this.right;
        }

        @Override
        public String toString() {
            return t+" "+left+" "+right;
        }


    }






    // After writing solution, quick scan for:
    //   array out of bounds
    //   special cases e.g. n=1?
    //
    // Big numbers arithmetic bugs:
    //   int overflow
    //   sorting, or taking max, or negative after MOD
    public static void main(String[] args)  throws IOException {
        Reader.init(System.in);
        BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
        int t= Reader.nextInt();
        long arr[]= new long[61];
        for (int i = 0; i <61 ; i++) {
            arr[i]= power(2,i+2, (long)3e18)-2;
        }
    //    System.out.println(Arrays.toString(arr));
        for (int tt = 0; tt <t ; tt++) {
            long n= Reader.nextLong();
            String str= Long.toBinaryString(n);
         //   System.out.println(str);
            if(n%2!=0){
                out.append(-1+"\n");
            }
            else{
                ArrayList<Integer>a= new ArrayList<>();
                while(n>=2){
                    for (int i = 60; i >=0 ; i--) {
                        if(arr[i]<=n){
                            a.add(1);
                            for (int j = 0; j <i ; j++) {
                               a.add(0);
                            }
                            n-=arr[i];
                            break;
                        }
                    }

                }
               out.append(a.size()+"\n");
                for (int i = 0; i <a.size() ; i++) {
                    out.append(a.get(i)+" ");
                }
                out.append("\n");
            }





        }

        out.flush();
        out.close();

    }





    static long modExp(long a, long b, long mod) {
        //System.out.println("a is " + a + " and b is " + b);
        if (a==1) return 1;
        long ans = 1;
        while (b!=0) {
            if (b%2==1) {
                ans = (ans*a)%mod;
            }
            a = (a*a)%mod;
            b/=2;
        }
        return ans;
    }


    public static long modmul(long a, long b, long mod) {
        return b == 0 ? 0 : ((modmul(a, b >> 1, mod) << 1) % mod + a * (b & 1)) % mod;
    }

    static long sum(long n){
        //   System.out.println("lol="+ (n*(n-1))/2);
        return (n*(n+1))/2;
    }






    public static ArrayList<Integer> Sieve(int n) {
        boolean arr[]= new boolean [n+1];
        Arrays.fill(arr,true);
        arr[0]=false;
        arr[1]=false;
        for (int i = 2; i*i <=n ; i++) {
            if(arr[i]){
                for (int j = 2; j <=n/i ; j++) {
                    int u= i*j;
                    arr[u]=false;
                }}
        }
        ArrayList<Integer> ans= new ArrayList<>();
        for (int i = 0; i <n+1 ; i++) {
            if(arr[i]){
                ans.add(i);
            }
        }
        return ans;
    }




    static long power( long x, long y, long p)
    {

        long res = 1;


        x = x % p;

        if (x == 0) return 0;

        while (y > 0)
        {

            if((y & 1)==1)
                res = (res * x) % p;


            y = y >> 1;
            x = (x * x) % p;
        }
        return res;
    }

    public static long ceil_div(long a, long b){
        return (a+b-1)/b;
    }

    static long gcd(long a, long b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

    static long lcm(long a, long b)
    {
        return (a*b)/gcd(a, b);
    }

}