Problem: Sasha likes investigating different math objects, for example, magic squares. But Sasha understands that magic squares have already been studied by hundreds of people, so he sees no sense of studying them further. Instead, he invented his own type of square — a prime square. 

A square of size n × n is called prime if the following three conditions are held simultaneously: 

  * all numbers on the square are non-negative integers not exceeding 10^5; 
  * there are no prime numbers in the square; 
  * sums of integers in each row and each column are prime numbers. 



Sasha has an integer n. He asks you to find any prime square of size n × n. Sasha is absolutely sure such squares exist, so just help him!

Input

The first line contains a single integer t (1 ≤ t ≤ 10) — the number of test cases.

Each of the next t lines contains a single integer n (2 ≤ n ≤ 100) — the required size of a square.

Output

For each test case print n lines, each containing n integers — the prime square you built. If there are multiple answers, print any.

Example

Input


2
4
2


Output


4 6 8 1
4 9 9 9
4 10 10 65
1 4 4 4
1 1
1 1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
void c_p_c() {}
class DisjointSet {
  unordered_map<int, int> parent;

 public:
  void makeSet(int N) {
    for (int i = 0; i < N; i++) parent[i] = i;
  }
  int Find(int v) {
    if (v == parent[v]) return v;
    return parent[v] = Find(parent[v]);
  }
  void Union(int a, int b) {
    int x = Find(a);
    int y = Find(b);
    parent[x] = y;
  }
};
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void PrimeFactors(long long n) {
  while (n % 2 == 0) {
    cout << 2 << " ";
    n = n / 2;
  }
  for (long long i = 3; i <= sqrt(n); i += 2) {
    while (n % i == 0) {
      cout << i << " ";
      n = n / i;
    }
  }
  if (n > 2) cout << n << " ";
}
template <typename T>
void parray(T a[], T n) {
  for (int i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << "\n";
}
vector<long long> uniqueelement(vector<long long> &a) {
  auto ip = unique((a).begin(), (a).end());
  a.resize(distance(a.begin(), ip));
  return a;
}
bool IsPrime(long long x) {
  if (x == 1) return false;
  for (int i = 2; 1ll * i * i <= x; i++) {
    if (x % i == 0) {
      return false;
    }
  }
  return true;
}
void leftrotate(string &s, int d) {
  reverse(s.begin(), s.begin() + d);
  reverse(s.begin() + d, s.end());
  reverse(s.begin(), s.end());
}
void rightrotate(string &s, int d) { leftrotate(s, s.length() - d); }
const int dx[4] = {-1, 1, 0, 0};
const int dy[4] = {0, 0, -1, 1};
int XX[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int YY[] = {-1, 0, 1, -1, 1, -1, 0, 1};
vector<long long> sieve() {
  vector<long long> primes;
  bool a[1000000] = {0};
  int mprimes = 1000000;
  a[0] = a[1] = true;
  for (int i = 2; i * i < mprimes; ++i) {
    if (a[i] == false) {
      for (int j = i * i; j < mprimes; j = j + i) {
        a[j] = true;
      }
    }
  }
  for (int i = 2; i < mprimes; ++i) {
    if (!a[i]) primes.push_back(i);
    ;
  }
  return primes;
}
void Ynot() {
  long long i, n, m, k, c = 0, sum = 0;
  bool flag = false;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      if (n % 2 == 1 and i == n - 1 and j == n / 2) {
        cout << 1 << " ";
      } else if (n % 2 == 1 and i == 0 and j == n / 2) {
        cout << 1 << " ";
      } else if (n % 2 == 1 and j == 0 and i == n / 2)
        cout << 1 << " ";
      else if (n % 2 == 1 and j == n - 1 and i == n / 2)
        cout << 1 << " ";
      else if (i == j)
        cout << 1 << " ";
      else if (i + j == n - 1)
        cout << 1 << " ";
      else
        cout << 0 << " ";
    }
    cout << "\n";
  }
}
int main() {
  c_p_c();
  int t = 1;
  cin >> t;
  for (int i = 0; i < t; i++) {
    Ynot();
  }
  return 0;
}