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 PYTHON3):

Solution:
''' ===============================
    -- @uthor : Kaleab Asfaw
    -- Handle : kaleabasfaw2010
    -- Bio    : High-School Student
    ==============================='''

# Fast IO
import sys
import os
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0
    def __init__(self, file): self._fd = file.fileno(); self.buffer = BytesIO(); self.writable = "x" in file.mode or "r" not in file.mode; self.write = self.buffer.write if self.writable else None
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b: break
            ptr = self.buffer.tell(); self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0; return self.buffer.read()
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)); self.newlines = b.count(b"\n") + (not b); ptr = self.buffer.tell(); self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1; return self.buffer.readline()
    def flush(self):
        if self.writable: os.write(self._fd, self.buffer.getvalue()); self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
    def __init__(self, file): self.buffer = FastIO(file); self.flush = self.buffer.flush; self.writable = self.buffer.writable; self.write = lambda s: self.buffer.write(s.encode("ascii")); self.read = lambda: self.buffer.read().decode("ascii"); self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout); input = lambda: sys.stdin.readline().rstrip("\r\n")
# Others
# from math import floor, ceil, gcd
# from decimal import Decimal as d
mod = 10**9+7
def lcm(x, y): return (x * y) / (gcd(x, y))
def fact(x, mod=mod):
    ans = 1
    for i in range(1, x+1): ans = (ans * i) % mod
    return ans
def arr2D(n, m, default=0): return [[default for j in range(m)] for i in range(n)]
def arr3D(n, m, r, default=0): return [[[default for k in range(r)] for j in range(m)] for i in range(n)]
def sortDictV(x): return {k: v for k, v in sorted(x.items(), key = lambda item : item[1])}
class DSU:
    def __init__(self, length): self.length = length; self.parent = [-1] * self.length  # O(log(n))
    def getParent(self, node, start):  # O(log(n))
        if node >= self.length: return False
        if self.parent[node] < 0:
            if start != node: self.parent[start] = node
            return node
        return self.getParent(self.parent[node], start)
    def union(self, node1, node2):  # O(log(n))
        parent1 = self.getParent(node1, node1); parent2 = self.getParent(node2, node2)
        if parent1 == parent2: return False
        elif self.parent[parent1] <= self.parent[parent2]: self.parent[parent1] += self.parent[parent2]; self.parent[parent2] = parent1
        else: self.parent[parent2] += self.parent[parent1]; self.parent[parent1] = parent2
        return True
    def getCount(self, node):  return -self.parent[self.getParent(node, node)] # O(log(n))

def SieveOfEratosthenes(n): 
    ans = []
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    for p in range(2, n+1): 
        if prime[p]: 
            ans.append(p)
    return ans

from bisect import bisect_left

prime = SieveOfEratosthenes(10**5) 
def findNum(n):
    ind = bisect_left(prime, n)
    # print("ind", ind, prime[ind-1: ind+2], prime[ind])
    for i in range(ind, len(prime)):
        ind1 = bisect_left(prime, prime[i] - n + 1)
        if prime[ind1] == prime[i] - n + 1:
            continue
        else:
            return prime[i] - n + 1

def solve(n):
    val = findNum(n)
    lst = arr2D(n, n, default=1)
    for i in range(n):
        lst[i][i] = val
    for i in lst:
        for j in i:
            print(j, end=" ")
        print()

for _ in range(int(input())): # Multicase
    n = int(input())
    solve(n)