Problem: A mad scientist Dr.Jubal has made a competitive programming task. Try to solve it!

You are given integers n,k. Construct a grid A with size n × n consisting of integers 0 and 1. The very important condition should be satisfied: the sum of all elements in the grid is exactly k. In other words, the number of 1 in the grid is equal to k.

Let's define:

  * A_{i,j} as the integer in the i-th row and the j-th column. 
  * R_i = A_{i,1}+A_{i,2}+...+A_{i,n} (for all 1 ≤ i ≤ n). 
  * C_j = A_{1,j}+A_{2,j}+...+A_{n,j} (for all 1 ≤ j ≤ n). 
  * In other words, R_i are row sums and C_j are column sums of the grid A. 
  * For the grid A let's define the value f(A) = (max(R)-min(R))^2 + (max(C)-min(C))^2 (here for an integer sequence X we define max(X) as the maximum value in X and min(X) as the minimum value in X). 



Find any grid A, which satisfies the following condition. Among such grids find any, for which the value f(A) is the minimum possible. Among such tables, you can find any.

Input

The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. Next t lines contain descriptions of test cases.

For each test case the only line contains two integers n, k (1 ≤ n ≤ 300, 0 ≤ k ≤ n^2).

It is guaranteed that the sum of n^2 for all test cases does not exceed 10^5.

Output

For each test case, firstly print the minimum possible value of f(A) among all tables, for which the condition is satisfied.

After that, print n lines contain n characters each. The j-th character in the i-th line should be equal to A_{i,j}.

If there are multiple answers you can print any.

Example

Input


4
2 2
3 8
1 0
4 16


Output


0
10
01
2
111
111
101
0
0
0
1111
1111
1111
1111

Note

In the first test case, the sum of all elements in the grid is equal to 2, so the condition is satisfied. R_1 = 1, R_2 = 1 and C_1 = 1, C_2 = 1. Then, f(A) = (1-1)^2 + (1-1)^2 = 0, which is the minimum possible value of f(A).

In the second test case, the sum of all elements in the grid is equal to 8, so the condition is satisfied. R_1 = 3, R_2 = 3, R_3 = 2 and C_1 = 3, C_2 = 2, C_3 = 3. Then, f(A) = (3-2)^2 + (3-2)^2 = 2. It can be proven, that it is the minimum possible value of f(A).

Solution (in JAVA):

Solution:


import com.sun.org.apache.xpath.internal.functions.FuncFalse;
import javafx.util.Pair;

import java.io.*;
import java.util.*;

import static java.lang.Math.*;

public class Contest {
    public static class pair implements Comparable<pair> {
        long l;
        long r;


        public pair(long lef, long ri) {
            l = lef;
            r = ri;
        }

        @Override
        public int compareTo(pair p) {
            if (l == p.l)
                return Long.compare(r, p.r);
            return Long.compare(l, p.l);
        }
    }

    static int n, m;
    static ArrayList<Integer>[] adj;
    static int dist[];
    static int parent[];
    static long inf = (long) 1e18;
    static int INF = (int) 1e9;


    public static boolean isSimilar(char[] Arr, int s, int k) {
        for (int i = s; i < k - 1; i++) {
            if (Arr[i + 1] != Arr[i])
                return false;
        }
        return true;
    }

    public static int ciel(int d, int mod) {
        if (d % mod == 0)
            return d / mod;
        return (d / mod) + 1;
    }


    public static long gcd(long u, long v) {
        if (v == 0)
            return u;
        return gcd(v, u % v);
    }

    static int k, Arr[], lefAcc[], riAcc[];
    static PrintWriter pw;

    static long memo[][];
    static char[][] Matrix;
    static int[] dx = new int[]{-1, 1, 0, 0};

    static int[] dy = new int[]{0, 0, -1, 1};


    static boolean valid(int i, int j) {

        return i != -1 && j != -1 && i != n && j != m;

    }


    static boolean[][] vis = new boolean[n][m];

    static void dfs2(int i, int j) {

        vis[i][j] = true;                    //mark as visited

        for (int k = 0; k < 4; ++k) {

            int x = i + dx[k], y = j + dy[k];

            if (valid(x, y) && Matrix[x][y] != '#' && !vis[x][y])

                dfs2(x, y);

        }

    }

    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        pw = new PrintWriter(System.out);

        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            int k = sc.nextInt();
            int[][] Matrix = new int[n][n];
            int c = 0;

            out:
            while (k > 0) {
                for (int i = c; i < n; i++) {
                    Matrix[i - c][i] = 1;
                    k--;
                    if (k <= 0) break out;
                }
                for (int i = n - c,j=0; i < n; j++,i++) {
                    Matrix[i ][j] = 1;
                    k--;
                    if (k <= 0) break out;
                }
                c++;
            }
            int lmin = 10000000;
            int lmax = 0;
            int cmax = 0;
            int cmin = 10000000;
            for (int i = 0; i < n; i++) {
                int sum = 0;
                for (int j = 0; j < n; j++)
                    sum += Matrix[i][j];
                lmax = max(lmax, sum);
                lmin = min(lmin, sum);
            }
            for (int i = 0; i < n; i++) {
                int sum = 0;
                for (int j = 0; j < n; j++)
                    sum += Matrix[j][i];
                cmax = max(cmax, sum);
                cmin = min(cmin, sum);
            }
            int dl = (lmax - lmin);
            int dc = (cmax - cmin);
            long ans = dl * dl + dc * dc;
            pw.println(ans);
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < n; j++)
                    pw.print(Matrix[i][j]);
                pw.println();
            }
        }
        pw.close();

    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

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

        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec)
                        f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

        public boolean ready() throws IOException, IOException {
            return br.ready();
        }

    }
}