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:
# cook your dish here
'''
#include <iostream>
#include<bits/stdc++.h>
#include<math.h>
using namespace std;

int main() {
	int t;
	scanf("%d",&t);
	while(t--)
	{
	    long long int x,y,r,l,i;
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void decToBinary(int n,vector<int>* v) 
{ 
    // array to store binary number 
    int binaryNum[50]; 
  
    // counter for binary array 
    int i = 0; 
    while (n > 0) { 
  
        // storing remainder in binary array 
        binaryNum[i] = n % 2;  



'''
import math
t=int(input())
for _ in range(0,t):
    n=input()
    lst=[]
    for i in range(len(n)-1,-1,-1):
        if n[i]!='0':
            lst.append(n[i]+'0'*(len(n)-1-i))
    print(len(lst))
    for i in range(0,len(lst)):
        print(lst[i],end=' ')
    print()