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:
def main():
    #from collections import defaultdict
    #from collections import Counter
    #import math
    # from fractions import Fraction
    # import heapq
    #from bisect import bisect_left
    import sys
    input = sys.stdin.buffer.readline
    #mod=998244353
    def GCD(x, y):
        while (y):
            x, y = y, x % y
        return x
    def lcm(a,b):
        return (a*b)//GCD(a,b)
    # Modified merge for inversion
    def merge(a,start,mid,end,d):
        p=start
        q=mid+1
        arr=[0]*(end-start+1)
        k=0
        coun=0
        for i in range(start,end+1):
            if p > mid:
                arr[k]=a[q]
                q+=1
                k+=1
            elif q > end:
                arr[k]=a[p]
                d[a[p]] += coun
                k+=1
                p+=1
            elif a[p] < a[q]:
                arr[k]=a[p]
                d[a[p]]+=coun
                k+=1
                p+=1
            else:
                arr[k]=a[q]
                k+=1
                q+=1
                coun+=1

        for i in range(k):
            a[start]=arr[i]
            start+=1
    def mergesort(a, start, end,d):
        if start<end:
            mid=(start+end)//2
            mergesort(a,start,mid,d)
            mergesort(a,mid+1,end,d)
            merge(a,start,mid,end,d)
    # inversion return dic for every index inversion count
    def inversion(l):
        d={}
        for i in l:
            d[i]=0
        mergesort(l,0,len(l)-1,d)
        return d
    def binary_search_lower(A,low,high,k):
        if high-low!=1:
            mid=(low+high)//2
            x = A[mid]
            if x>=k:
                return binary_search_lower(A,low,mid,k)
            else:
                return binary_search_lower(A,mid,high,k)
        if A[low]>=k:
            return low
        else:
            return high
    # Function to find power
    def binary_expo(x,n):
        if n==0:
            return 1
        elif n%2==0:
            return binary_expo(x*x,n//2)
        else:
            return x*binary_expo(x*x,(n-1)//2)
    # Function to find power with given modulus
    def modu_expo(x,n,m):
        if n==0:
            return 1
        elif n%2==0:
            return modu_expo((x*x)%m,n//2,m)
        else:
            return (x*modu_expo((x*x)%m,(n-1)//2,m))%m
    # Function for modular multiplicative inverse via using extended_GCD_Euclid's
    def eEuclid(c, m):
        if m == 0:
            global d
            global x
            global y
            d = c
            x = 1
            y = 0
        else:
            eEuclid(m, c % m)
            temp = x
            x = y
            y = temp - (c // m) * y
    def mod_Inv(c, m):
        eEuclid(c, m)
        return (x % m + m) % m
    # Function for nCr%p using fermat (p is prime)
    def ncr(n, r, p):
        num = den = 1
        for i in range(r):
            num = (num * (n - i)) % p
            den = (den * (i + 1)) % p
        return (num * pow(den,
                          p - 2, p)) % p
    #*************************************************************
    for _ in range(int(input())):
        n=int(input())
        ans=[]
        i=1
        while n>10:
            x=(n%10)
            if x!=0:
                ans.append(x*i)
            i = i * 10
            n=n//10
        ans.append(n * i)
        print(len(ans))
        for i in ans:
            print(i,end=" ")
        print()

if __name__ == "__main__":
    main()