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>
#pragma GCC optimize("O3")
using namespace std;
template <class T>
using pq = priority_queue<T>;
template <class T>
using pqg = priority_queue<T, vector<T>, greater<T>>;
int scan() { return getchar(); }
template <class T>
void scan(T a) {
  cin >> a;
}
void scan(int &a) { cin >> a; }
void scan(long long &a) { cin >> a; }
void scan(char &a) { cin >> a; }
void scan(double &a) { cin >> a; }
void scan(long double &a) { cin >> a; }
void scan(char a[]) { scanf("%s", a); }
void scan(string &a) { cin >> a; }
template <class T>
void scan(vector<T> &);
template <class T, size_t size>
void scan(array<T, size> &);
template <class T, class L>
void scan(pair<T, L> &);
template <class T, size_t size>
void scan(T (&)[size]);
template <class T>
void scan(vector<T> &a) {
  for (auto &i : a) scan(i);
}
template <class T>
void scan(deque<T> &a) {
  for (auto &i : a) scan(i);
}
template <class T, size_t size>
void scan(array<T, size> &a) {
  for (auto &i : a) scan(i);
}
template <class T, class L>
void scan(pair<T, L> &p) {
  scan(p.first);
  scan(p.second);
}
template <class T, size_t size>
void scan(T (&a)[size]) {
  for (auto &i : a) scan(i);
}
template <class T>
void scan(T &a) {
  cin >> a;
}
void IN() {}
template <class Head, class... Tail>
void IN(Head &head, Tail &...tail) {
  scan(head);
  IN(tail...);
}
string stin() {
  string s;
  cin >> s;
  return s;
}
template <class T, class S>
inline bool chmax(T &a, S b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T, class S>
inline bool chmin(T &a, S b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
vector<int> iota(int n) {
  vector<int> a(n);
  iota(begin(a), end(a), 0);
  return a;
}
template <class T>
void UNIQUE(vector<T> &x) {
  sort(begin(x), end(x));
  x.erase(unique(begin(x), end(x)), x.end());
}
int in() {
  int x;
  cin >> x;
  return x;
}
long long lin() {
  unsigned long long x;
  cin >> x;
  return x;
}
void print() { putchar(' '); }
void print(bool a) { cout << a; }
void print(int a) { cout << a; }
void print(long long a) { cout << a; }
void print(char a) { cout << a; }
void print(string &a) { cout << a; }
void print(double a) { cout << a; }
template <class T>
void print(const vector<T> &);
template <class T, size_t size>
void print(const array<T, size> &);
template <class T, class L>
void print(const pair<T, L> &p);
template <class T, size_t size>
void print(const T (&)[size]);
template <class T>
void print(const vector<T> &a) {
  if (a.empty()) return;
  print(a[0]);
  for (auto i = a.begin(); ++i != a.end();) {
    cout << " ";
    print(*i);
  }
  cout << endl;
}
template <class T>
void print(const deque<T> &a) {
  if (a.empty()) return;
  print(a[0]);
  for (auto i = a.begin(); ++i != a.end();) {
    cout << " ";
    print(*i);
  }
}
template <class T, size_t size>
void print(const array<T, size> &a) {
  print(a[0]);
  for (auto i = a.begin(); ++i != a.end();) {
    cout << " ";
    print(*i);
  }
}
template <class T, class L>
void print(const pair<T, L> &p) {
  cout << '(';
  print(p.first);
  cout << ",";
  print(p.second);
  cout << ')';
}
template <class T>
void print(set<T> &x) {
  for (auto e : x) print(e), cout << " ";
  cout << endl;
}
template <class T>
void print(multiset<T> &x) {
  for (auto e : x) print(e), cout << " ";
  cout << endl;
}
template <class T, size_t size>
void print(const T (&a)[size]) {
  print(a[0]);
  for (auto i = a; ++i != end(a);) {
    cout << " ";
    print(*i);
  }
}
template <class T>
void print(const T &a) {
  cout << a;
}
int out() {
  putchar('\n');
  return 0;
}
template <class T>
int out(const T &t) {
  print(t);
  putchar('\n');
  return 0;
}
template <class Head, class... Tail>
int out(const Head &head, const Tail &...tail) {
  print(head);
  putchar(' ');
  out(tail...);
  return 0;
}
long long gcd(long long a, long long b) {
  while (b) {
    long long c = b;
    b = a % b;
    a = c;
  }
  return a;
}
long long lcm(long long a, long long b) {
  if (!a || !b) return 0;
  return a * b / gcd(a, b);
}
vector<pair<long long, long long>> factor(long long x) {
  vector<pair<long long, long long>> ans;
  for (long long i = 2; i * i <= x; i++)
    if (x % i == 0) {
      ans.push_back({i, 1});
      while ((x /= i) % i == 0) ans.back().second++;
    }
  if (x != 1) ans.push_back({x, 1});
  return ans;
}
vector<int> divisor(int x) {
  vector<int> ans;
  for (int i = 1; i * i <= x; i++)
    if (x % i == 0) {
      ans.push_back(i);
      if (i * i != x) ans.push_back(x / i);
    }
  return ans;
}
template <typename T>
void zip(vector<T> &x) {
  vector<T> y = x;
  sort(begin(y), end(y));
  for (int i = 0; i < x.size(); ++i) {
    x[i] = distance((y).begin(), lower_bound(begin(y), end(y), (x[i])));
  }
}
int popcount(long long x) { return __builtin_popcountll(x); }
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int n) { return uniform_int_distribution<int>(0, n - 1)(rng); }
template <typename T>
void shuffle(vector<T> &v) {
  for (long long i = v.size() - 1; i >= 1; --i) {
    swap(v[i], v[rnd(i)]);
  }
}
vector<string> YES{"NO", "YES"};
vector<string> Yes{"No", "Yes"};
vector<string> yes{"no", "yes"};
template <class... T>
void err(const T &...) {}
template <typename T>
struct edge {
  int from, to;
  T cost;
  int id;
  edge(int to, T cost) : from(-1), to(to), cost(cost) {}
  edge(int from, int to, T cost) : from(from), to(to), cost(cost) {}
  edge &operator=(const int &x) {
    to = x;
    return *this;
  }
  operator int() const { return to; }
};
template <typename T>
using Edges = vector<edge<T>>;
template <typename T>
using WeightedTree = vector<Edges<T>>;
template <typename T>
using Edges = vector<edge<T>>;
template <typename T>
using WeightedGraph = vector<Edges<T>>;
template <typename T>
using Matrix = vector<vector<T>>;
using Graph = vector<vector<int>>;
using tree = vector<vector<int>>;
tree getTree(int n, int offset = 1) {
  tree res(n);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a -= offset, b -= offset;
    res[a].emplace_back(b);
    res[b].emplace_back(a);
  }
  return res;
}
template <typename T>
WeightedTree<T> getWeightedTree(int n, int offset = 1) {
  WeightedTree<T> res(n);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a -= offset, b -= offset;
    T c;
    cin >> c;
    res[a].emplace_back(b, c);
    res[b].emplace_back(a, c);
  }
  return res;
}
Graph getGraph(int n, int m, int offset = 1) {
  Graph res(n);
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    a -= offset, b -= offset;
    res[a].emplace_back(b);
    res[b].emplace_back(a);
  }
  return res;
}
template <typename T>
WeightedGraph<T> getWeightedGraph(int n, int m, int offset = 1) {
  WeightedGraph<T> res(n);
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    T c;
    cin >> c;
    res[a].emplace_back(b, c);
    res[b].emplace_back(a, c);
  }
}
struct Setup_io {
  Setup_io() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cout << fixed << setprecision(15);
  }
} setup_io;
int main() {
  int T;
  IN(T);
  while (T--) {
    int n, k;
    IN(n, k);
    vector<string> ans(n, string(n, '0'));
    for (long long i = 0; i < n; ++i) {
      for (long long j = 0; j < n; ++j) {
        if (k) {
          ans[(i + j) % n][j] = '1';
          k--;
        }
      }
    }
    vector<int> v, w;
    for (long long i = 0; i < n; ++i) {
      v.emplace_back(0), w.emplace_back(0);
      for (long long j = 0; j < n; ++j)
        v.back() += ans[i][j] - '0', w.back() += ans[j][i] - '0';
    }
    sort(begin(v), end(v)), sort(begin(w), end(w));
    cout << (v.back() - v.front()) * (v.back() - v.front()) +
                (w.back() - w.front()) * (w.back() - w.front())
         << '\n';
    for (long long i = 0; i < n; ++i) cout << ans[i] << '\n';
  }
}