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 collections import Counter,defaultdict,deque
#import heapq as hq
#import itertools
from operator import itemgetter
#from itertools import count, islice
#from functools import reduce
#alph = 'abcdefghijklmnopqrstuvwxyz'
#from math import factorial as fact
#a,b = [int(x) for x in input().split()]
#sarr = [x for x in input().strip().split()]
import math
import sys
input=sys.stdin.readline
#sys.setrecursionlimit(2**30)

def solve():
    #n,k = [int(x) for x in input().split()]
    #print(k//n+k)
    n = int(input())
    res = []
    p = 0
    while n>0:
        d = (n%10)
        if d!=0:
            res.append(d*10**p)
        n//=10
        p+=1
    print(len(res))
    print(*res)
    
tt = int(input())
for test in range(tt):
    solve()















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