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 $n \times n$ 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 $1 \le i \le n$).  $C_j = A_{1,j}+A_{2,j}+...+A_{n,j}$ (for all $1 \le j \le n$).  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 $t$ ($1 \le t \le 100$) — the number of test cases. Next $t$ lines contain descriptions of test cases.

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

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 $f(A)$.

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = "import sys\ninput = sys.stdin.readline\n\nimport math\n\nT = int(input())\n\n\ndef f(M):\n    minR = math.inf\n    maxR = -math.inf\n    minC = math.inf\n    maxC = -math.inf\n\n    for i in range(len(M)):\n        sumRow = sum(M[i])\n        minR = min(minR, sumRow)\n        maxR = max(maxR, sumRow)\n\n        sumCol = sum([M[el][i] for el in range(len(M))])\n        maxC = max(maxC, sumCol)\n        minC = min(minC, sumCol)\n\n    return (maxR - minR)**2 + (maxC - minC)**2\n\nfor t in range(T):\n    N, K = [int(_) for _ in input().split()]\n    M = [[0] * N for i in range(N)]\n\n    # save = set()\n\n    for i in range(K):\n        # assert (i%N, (i//N + i)%N) not in save\n        # save.add((i%N, (i//N + i)%N))\n        M[i%N][(i//N + i)%N] = 1\n\n    print(f(M))\n    for row in M:\n        print(''.join(map(str, row)))\n"
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13267