human
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 
"""Te
Write me a Python code to print t answers to the test cases?
assistant
#Here is Python code to print t answers to the test cases.
mplate for Python Competitive Programmers prepared by pa.n.ik and Kabeer seth """

# to use the print and division function of Python3
from __future__ import division, print_function

"""value of mod"""
MOD = 998244353
mod = 10**9 + 7

"""use resource"""
# import resource
# resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])

"""for factorial"""

# def prepare_factorial():
#     fact = [1]
#     for i in range(1, 2005):
#         fact.append((fact[-1] * i) % mod)
#     ifact = [0] * 2005
#     ifact[2004] = pow(fact[2004], mod - 2, mod)
#     for i in range(2004, 0, -1):
#         ifact[i - 1] = (i * ifact[i]) % mod
#
#     return fact, ifact

"""uncomment next 4 lines while doing recursion based question"""
# import threading
# threading.stack_size(1<<27)
import sys
# sys.setrecursionlimit(10000)

"""uncomment modules according to your need"""
from bisect import bisect_left, bisect_right, insort
# from itertools import repeat
# from math import floor, ceil, sqrt, degrees, atan, pi, log
from heapq import heappop, heapify, heappush
# from random import randint as rn
# from Queue import Queue as Q
# from collections import Counter, defaultdict, deque
# from copy import deepcopy
'''
def modinv(n, p):
    return pow(n, p - 2, p)
'''

#
# def ncr(n, r,  fact, ifact):  # for using this uncomment the lines calculating fact and ifact
#     t = (fact[n] * (ifact[r]*ifact[n-r]) % mod)% mod
#     return t


def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def input(): return sys.stdin.readline().strip()

"""*****************************************************************************************"""

# def GCD(x, y):
#     while (y):
#         x, y = y, x % y
#     return x
#
# def lcm(x, y):
#     return (x*y)//(GCD(x, y))

# def get_xor(n):
#     return [n,1,n+1,0][n%4]

# def bin_pow(a, b):
#     res = 1
#     while b:
#         if b&1:
#             res = (res * a)
#             b -= 1
#   ...