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 
from collections import defaultdict as dc
from heapq import *
import math
from bisect import bisect_left,bisect
#bisect gives x and p[x] is element greater than it and out of bound for last one 
#p[x-1] gives equal or smaller and no error for any element.
import sys
from collections import deque as dq
from heapq import heapify,heappush,heappop
mod=10**9 +7
def sinp():
    p=str(input())
    return p
def inp(): 
    p=int(input())
    return p
def line():
    p=list(map(int,input().split()))
    return p
def read_mat():
    n=inp()
    a=[]
    for i in range(n):
        a.append(line())
    return a
def digit(n):
    s=str(n)
    p=0
    for i in s:
        p+=(int(i))**2
    return p
def check(p,i,j):
    z=i+(j-i)//2
    if p[z]-p[i] == p[j]-p[z]:
        #print(p[z]-p[i],p[j]-p[z])
        return j-i
    return 0
def solve(a):
    z=0
    ans=[]
    l=str(a)
    r=1
    for i in range(len(l)-1,-1,-1):
        p=int(l[i])*r
        r=r*10
        # print(p)
        if p>0:
            ans.append(p)
    return ans
    
for test in range(inp()):
    a=inp()
    # n=len(a)
    l=solve(a)
    print(len(l))
    print(*l)