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:
#input template
from sys import stdin, stdout
from collections import Counter, OrderedDict
import math
cin = stdin.readline
cout = stdout.write
mp = lambda: list(map(int, cin().split()))

def chars(): #function for taking string input as character array since string in python is immutable
    s = cin()
    return(list(s[:len(s) - 1]))

#print list	
def pl(a):	
	for val in a:
		cout(str(val) + ' ')
	cout('\n')
		
#main

t, = mp()
for _ in range(t):
	s = chars()
	
	n = len(s)
	count = 0
	l = ''
	for i in range(n):
		if s[i] != '0':
			count += 1
			l += s[i] + '0'*(n-i-1) + ' '
			
	cout(str(count) + '\n' + l + '\n')