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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int prime[1000000];
vector<int> pr;
void calculateprime() {
  prime[0] = prime[1] = 1;
  for (int i = 2; i <= 1000000; i++) {
    if (prime[i] == 0) {
      pr.push_back(i);
      for (long long j = i * 2; j <= 1000000; j += i) {
        prime[j] = 1;
      }
    }
  }
}
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
bool isprime(long long n) {
  for (long long i = 2; (long long)i * i <= n; i++) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
long long fib(int n) {
  long long a = 0, b = 1, c, i;
  if (n == 0) return a;
  for (i = 2; i <= n; i++) {
    c = (a + b);
    a = b;
    b = c;
  }
  return b;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long extendedgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long x1, y1;
  long long d = extendedgcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return d;
}
long long findgrouplcm(vector<int> &arr, int n) {
  long long ans = arr[0];
  for (long long i = 1; i < n; i++) ans = (arr[i] * (ans / (gcd(arr[i], ans))));
  return ans;
}
long long powerwithmod(long long x, unsigned long long y, long long p) {
  long long res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
bool comp(const pair<long long, long long> &a,
          const pair<long long, long long> &b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first > b.first;
  return false;
}
void solve() {
  int n, k, fun;
  cin >> (n) >> k;
  if (k % n == 0) {
    fun = 0;
  } else {
    fun = 2;
  }
  cout << fun << "\n";
  int arr[n][n];
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < n; j++) arr[i][j] = 0;
  int pos[n];
  for (long long i = 0; i < n; i++) pos[i] = i;
  while (k) {
    for (int i = 0; i < n and k; i++) {
      arr[i][pos[i]] = 1;
      pos[i] = (pos[i] + 1) % n;
      --k;
    }
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) cout << arr[i][j];
    cout << "\n";
  }
}
int main() {
  fast();
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}