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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1 << 29;
const long long MOD = 1000000007;
const long long FMOD = 998244353;
const long long MAX = 10000000005;
const long long MIN = -10000000005;
int cmp(const void* a, const void* b);
long long maxv(long long a, long long b);
long long minv(long long a, long long b);
long long gcd(long long u, long long v);
long long digits(long long n);
bool ispoweroftwo(long long n);
bool isvowel(char x);
long long chartoint(char ch);
long long CEIL(long long x, long long y);
long long FLOOR(long long x, long long y);
int main() {
  long long t;
  scanf("%lld", &t);
  while (t--) {
    long long n;
    scanf("%lld", &n);
    long long p = 1;
    vector<long long> ans;
    while (n) {
      long long x = n % 10;
      if (x != 0) {
        x = x * p;
        ans.push_back(x);
      }
      n = n / 10;
      p = p * 10;
    }
    long long x1 = ans.size();
    printf("%lld\n", x1);
    for (long long i = 0; i < (x1); ++i) printf("%lld ", ans[i]);
    printf("\n");
  }
  return 0;
}
int cmp(const void* a, const void* b) {
  if (*(long long*)a - *(long long*)b < 0) return -1;
  if (*(long long*)a - *(long long*)b > 0) return 1;
  return 0;
}
long long maxv(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long minv(long long a, long long b) {
  if (a < b) return a;
  return b;
}
long long gcd(long long u, long long v) {
  if (v == 0) return u;
  return gcd(v, u % v);
}
long long digits(long long n) { return floor(log10(n)) + 1; }
bool ispoweroftwo(long long x) { return x && (!(x & (x - 1))); }
bool isvowel(char x) {
  return (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u');
}
long long chartoint(char ch) {
  if (ch >= 'A' && ch <= 'Z')
    return (ch - 'A');
  else if (ch >= '0' && ch <= '9')
    return (ch - '0');
  else if (ch >= 'a' && ch <= 'z')
    return (ch - 'a');
  else
    return 0;
}
long long CEIL(long long x, long long y) {
  if (x % y == 0)
    return (x / y);
  else
    return (x / y + 1);
}
long long FLOOR(long long x, long long y) {
  if (x % y == 0)
    return (x / y);
  else
    return (x / y - 1);
}