Problem: A positive (strictly greater than zero) integer is called round if it is of the form d00...0. In other words, a positive integer is round if all its digits except the leftmost (most significant) are equal to zero. In particular, all numbers from 1 to 9 (inclusive) are round.

For example, the following numbers are round: 4000, 1, 9, 800, 90. The following numbers are not round: 110, 707, 222, 1001.

You are given a positive integer n (1 ≤ n ≤ 10^4). Represent the number n as a sum of round numbers using the minimum number of summands (addends). In other words, you need to represent the given number n as a sum of the least number of terms, each of which is a round number.

Input

The first line contains an integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input. Then t test cases follow.

Each test case is a line containing an integer n (1 ≤ n ≤ 10^4).

Output

Print t answers to the test cases. Each answer must begin with an integer k — the minimum number of summands. Next, k terms must follow, each of which is a round number, and their sum is n. The terms can be printed in any order. If there are several answers, print any of them.

Example

Input


5
5009
7
9876
10000
10


Output


2
5000 9
1
7 
4
800 70 6 9000 
1
10000 
1
10 

Solution (in PYTHON3):

Solution:
from sys import stdin, stdout
from collections import deque

rl = lambda: stdin.readline()
rll = lambda: stdin.readline().split()
rli = lambda: map(int, stdin.readline().split())

def toL(num):
	ans = deque()
	while num:
		num, dig = divmod(num, 10)
		ans.appendleft(dig)
	return ans

def main():
	T = int(rl())
	for _ in range(T):
		n = int(rl())
		L = toL(n)
		k = 0
		ans = []
		for i, num in enumerate(L):
			if num == 0: continue
			k += 1
			bi = len(L) - i - 1
			ans.append(num * 10**(bi))
		stdout.write(str(k));stdout.write("\n")
		ansstr = " ".join(str(pt) for pt in ans)
		stdout.write(ansstr);stdout.write("\n")
	stdout.close()

if __name__ == "__main__":
	main()