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:
t=int(input())
while(t>0):
    n=int(input())
    s=len(str(n))
    c=0
    l=[]
    g=s-1
    if(n%10**(s-1)==0 or s==1):
        print(1)
        print(n)
    else:
        for i in range(g):
            
            f=n%(10**(s-1))
            n=n//(10**(s-1))
            a=len(str(f))
            l.append(n*10**(s-1))

        #print(n)
        #print(f)
            
            n=f
            #print(n)
            
            #print(a)
            if(f%(10**(a-1))==0) or len(str(f))==1:
                c+=1
                l.append(f)
                c+=1
                #print(l)
                break
            else:
                c+=1
                s=a
        print(c)   
        print(' '.join(map(str,l)))
    t-=1