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:
"""
    Author: Sagar Pandey

"""
# ---------------------------------------------------Import Libraries---------------------------------------------------
import sys
import os
from math import sqrt, log, log2, log10, gcd, floor, pow, sin, cos, tan, pi, inf, factorial
from copy import copy, deepcopy
from sys import exit, stdin, stdout
from collections import Counter, defaultdict, deque
from itertools import permutations
import heapq
from bisect import bisect_left as bl
# If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect

# If the element is already present in the list,
# the right most position where element has to be inserted is r

# ---------------------------------------------------Global Variables---------------------------------------------------
# sys.setrecursionlimit(100000000)
mod = 1000000007
# ---------------------------------------------------Helper Functions---------------------------------------------------
iinp = lambda: int(sys.stdin.readline())
inp = lambda: sys.stdin.readline().strip()
strl = lambda: list(inp().strip().split(" "))
intl = lambda: list(map(int, inp().split(" ")))
mint = lambda: map(int, inp().split())
flol = lambda: list(map(float, inp().split(" ")))
flush = lambda: stdout.flush()


def permute(nums):
    def fun(arr, nums, cur, v):
        if len(cur) == len(nums):
            arr.append(cur.copy())
        i = 0
        while i < len(nums):
            if v[i]:
                i += 1
                continue
            else:
                cur.append(nums[i])
                v[i] = 1
                fun(arr, nums, cur, v)
                cur.pop()
                v[i] = 0
                i += 1
            # while i<len(nums) and nums[i]==nums[i-1]:i+=1    # Uncomment for unique permutations
        return arr

    res = []
    nums.sort()
    v = [0] * len(nums)
    return fun(res, nums, [], v)


def subsets(res, index, arr, cur):
    res.append(cur.copy())
    for i in range(index, len(arr)):
        cur.append(arr[i])
        subsets(res, i + 1, arr, cur)
        cur.pop()
    return res


def sieve(N):
    root = int(sqrt(N))
    primes = [1] * (N + 1)
    primes[0], primes[1] = 0, 0
    for i in range(2, root + 1):
        if primes[i]:
            for j in range(i * i, N + 1, i):
                primes[j] = 0
    return primes


def bs(arr, l, r, x):
    if x < arr[0] or x > arr[len(arr) - 1]:
        return -1
    while l <= r:
        mid = l + (r - l) // 2
        if arr[mid] == x:
            return mid
        elif arr[mid] < x:
            l = mid + 1
        else:
            r = mid - 1
    return -1


def isPrime(n):
    if n <= 1: return False
    if n <= 3: return True
    if n % 2 == 0 or n % 3 == 0: return False
    p = int(sqrt(n))
    for i in range(5, p + 1, 6):
        if n % i == 0 or n % (i + 2) == 0:
            return False
    return True


# -------------------------------------------------------Functions------------------------------------------------------

def solve():
    s=inp()
    n=len(s)
    p=0
    ans=[]
    for i in range(n-1,-1,-1):
        if int(s[i])!=0:
            ans.append(s[i]+"0"*p)
        p+=1
    print(len(ans))
    print(*ans[::-1])





# -------------------------------------------------------Main Code------------------------------------------------------
for _ in range(iinp()):
    solve()