Given a positive integer n. Find three distinct positive integers a, b, c such that a + b + c = n and \operatorname{gcd}(a, b) = c, where \operatorname{gcd}(x, y) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers x and y.

Input

The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Description of the test cases follows.

The first and only line of each test case contains a single integer n (10 ≤ n ≤ 10^9).

Output

For each test case, output three distinct positive integers a, b, c satisfying the requirements. If there are multiple solutions, you can print any. We can show that an answer always exists.

Example

Input


6
18
63
73
91
438
122690412


Output


6 9 3
21 39 3
29 43 1
49 35 7
146 219 73
28622 122661788 2

Note

In the first test case, 6 + 9 + 3 = 18 and \operatorname{gcd}(6, 9) = 3.

In the second test case, 21 + 39 + 3 = 63 and \operatorname{gcd}(21, 39) = 3.

In the third test case, 29 + 43 + 1 = 73 and \operatorname{gcd}(29, 43) = 1.
1
669278611
 10
10
52
91
23
44
99
28
54
99
51

334639303 334639307 1
 4 5 1
25 26 1
43 47 1
9 13 1
21 22 1
47 51 1
13 14 1
26 27 1
47 51 1
23 27 1

import os
import sys
from bisect import bisect, bisect_left, insort, insort_left
from heapq import heapify, heappop, heappush, nlargest, nsmallest
from io import BytesIO, IOBase
from math import inf, sqrt
from types import GeneratorType

# Fast IO for Python

BUFSIZE: int = 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")


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def input(): return sys.stdin.readline().rstrip("\r\n")


# Bootstrap used for recursion, use yield instead of return

def bootstrap(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
    return wrappedfunc


# Never use += for string concatenation
# Always use .join for string concatenation
# Use PyPy3 for Codeforces
# Use Python3 for Codechef

# Function to compute inverse of a under modulo m

def modInverse(a, m):

    g = gcd(a, m)

    if (g != 1):
        return -1

    else:
        # If a and m are relatively prime,
        # then modulo inverse is a^(m-2) mode m
        return power(a, m - 2, m)


# Function to compute x^y under modulo m

def power(x, y, m):

    if (y == 0):
        return 1

    p = power(x, y // 2, m) % m
    p = (p * p) % m

    if(y % 2 == 0):
        return p
    else:
        return ((x * p) % m)


# Function to return gcd of a and b

def gcd(a, b):
    if (a == 0):
        return b

    return gcd(b % a, a)


# Sieve of Eratosthenes to get prime upto n

def SieveOfEratosthenes(n):
    prime = [True for i in range(n+1)]
    p = 2
    while (p * p <= n):

        # If prime[p] is not
        # changed, then it is a prime
        if (prime[p] == True):

            # Update all multiples of p
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1

    ans = set()
    for p in range(2, n+1):
        if prime[p]:
            ans.add(p)

    return ans


MOD = 1000000007
INT_MAX = inf
INT_MIN = -inf


# Main function

def main():
    t = int(input())
    for _ in range(t):
        temp = list(map(int, input().split()))
        n = temp[0]
        for i in range(2, n):
            a = i
            b = n-1-a
            if gcd(a, b) == 1:
                print(' '.join(map(str, [a, b, 1])))
                break


main()
