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.util.*;
public class check4  {

    public static long count_pow(long no)
    {
        long ans=0;
        while(no>=2)
        {
            ans+=1;
            no/=2;

        }
        return ans;

    }

    public static void find1(long k,List<Long> al)
    {
        if(k==0)
        {
            return;
        }
        long k1=k+2;
        long t1= count_pow(k1);

        long k2=k1-(long)Math.pow(2,t1);

        al.add(t1-2);

        find1(k2,al);

    }
    public static StringBuilder find(long k)
    {
        List<Long> al=new ArrayList<>();
        find1(k-2,al);
        StringBuilder sb = new StringBuilder();
        sb.append(1);

        for(int i=0;i<al.size();i++)
        {
            long t1=al.get(i);
            while(t1>0)
            {
                sb.append('0');
                t1--;
            }
            sb.append('1');
        }
        return sb;

    }
    public static void main(String[] args) throws IOException{

        Reader sc=new Reader();
        PrintWriter out = new PrintWriter(System.out);
        int t=sc.nextInt();

        while(t-->0)
        {
            long k=sc.nextLong();
            if(k%2==1)
            {
                out.println(-1);
                continue;
            }
            StringBuilder sb = find(k);

            out.println(sb.length());
            for(int i=0;i<sb.length();i++)
            {
                out.print(sb.charAt(i)+" ");
            }
            out.println();

        }
        out.flush();
    }

    public static long power(long x,long power)
    {
        if(power==0) return 1;

        long ans1 = power(x,power/2);
        ans1=ans1*ans1;

        if(power%2==0) return ans1;
        return x*ans1;
    }

    //    (a*a_inverse) = 1(mod m)
//    finding a_inverse
    public static long modInverse(long a,long m)
    {
//        works only when m is prime
        long g=gcd(a,m);
        if(g==1) return power(a,m-2,m);
        else return -1;
    }

    //    (x^power)mod(m)
    public static long power(long x,long power,long m)
    {
        if(power==0) return 1;

        long p = power(x,power/2,m)%m;
        p = (p*p)%m;

        if(power%2==0) return p;
        return (x*p)%m;
    }
    public static long gcd(long small,long large)
    {
        if(small==0) return large;
        return gcd(large % small,small);
    }

    public static boolean isprime(long no)
    {
        if(no<=1) return false;
        if(no<=3) return true;
        if(no%2==0 || no%3==0) return false;
        // 6k+1 6k+5 can be prime
        for(long i=5;i*i<=no;i+=6)
        {
            if(no%(i)==0 || no%(i+2)==0)
                return false;
        }
        return true;
    }

    //    prime no smaller than or equal to n
    public static boolean[] prime_nos_till_no(int no)
    {
        boolean prime[]=new boolean[no+1];
//        prime[i]== true means prime
//        prime[i]== false means not a prime
//        initialize prime array as true

        Arrays.fill(prime,true);
        prime[0]=false;
        for(int i=2;i*i<=no;i++)
        {
            if(prime[i]==true)
            {
                for(int j=i*i;j<=no;j+=i)
                {
                    prime[j]=false;
                }
            }
        }
        return prime;

    }
    public static void shufflearray(long arr[])
    {


        int n=arr.length;
        Random rand=new Random();
        for(int i=0;i<n;i++)
        {
            long temp=arr[i];
            int randomPos=i+rand.nextInt(n-i);
            arr[i]=arr[randomPos];
            arr[randomPos]=temp;
        }
    }
    //       Arrays.sort(arr, new Comparator<pair>() {
//        //@Override
//        public int compare(pair o1, pair o2) {
//            long l1=o1.a-o2.a;
//            if(l1<0L) return -1;
//            if(l1==0) return 0;
//            return 1;
//        }
//    });
    static class Reader {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        public Reader() {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public Reader(String file_name) throws IOException {
            din = new DataInputStream(new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public String nextLine() throws IOException {
            byte[] buf = new byte[64]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n')
                    break;
                buf[cnt++] = (byte) c;
            }
            return new String(buf, 0, cnt);
        }
        public boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public String next() throws IOException{
            int c = read();
            while (isSpaceChar(c)) c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isSpaceChar(c));
            return res.toString();
        }


        public int nextInt() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');

            if (neg)
                return -ret;
            return ret;
        }

        public long nextLong() throws IOException {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }

        public double nextDouble() throws IOException {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();

            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');

            if (c == '.') {
                while ((c = read()) >= '0' && c <= '9') {
                    ret += (c - '0') / (div *= 10);
                }
            }

            if (neg)
                return -ret;
            return ret;
        }

        private void fillBuffer() throws IOException {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }

        private byte read() throws IOException {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }

        public void close() throws IOException {
            if (din == null)
                return;
            din.close();
        }

    }

}