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;
int main() {
  int t, i, j, m, n, p, r, s;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    if (n < 10) {
      printf("1\n%d\n", n);
    } else if (n < 100 && n % 10 == 0) {
      printf("1\n%d\n", n);
    } else if (n >= 100 && n < 1000 && n % 100 == 0) {
      printf("1\n%d\n", n);
    } else if (n > 10 && n < 100) {
      printf("2\n%d %d\n", (n / 10) * 10, n % 10);
    } else if (n > 100 && n < 1000) {
      s = n % 100;
      if (s < 10 || s % 10 == 0) {
        printf("2\n%d %d\n", (n / 100) * 100, s);
      } else
        printf("3\n%d %d %d\n", (n / 100) * 100, (s / 10) * 10, s % 10);
    } else if (n >= 1000) {
      m = n / 1000;
      p = n % 1000;
      r = p % 100;
      if (p == 0) {
        printf("1\n%d\n", n);
      } else if (p < 10) {
        printf("2\n%d %d\n", m * 1000, p);
      } else if (p >= 10 && p <= 100) {
        if (p >= 10 && p % 10 == 0) {
          printf("2\n%d %d\n", m * 1000, p);
        } else if (p % 100 == 0) {
          printf("2\n%d %d\n", m * 1000, p);
        } else
          printf("3\n%d %d %d\n", m * 1000, p - (p % 10), n % 10);
      } else if (p > 100 && p < 1000) {
        if (p % 100 == 0) {
          printf("2\n%d %d\n", m * 1000, p);
        } else if (r <= 10) {
          printf("3\n%d %d %d\n", (n / 1000) * 1000, (p / 100) * 100, r);
        } else if (r > 10 && r % 10 == 0) {
          printf("3\n%d %d %d\n", (n / 1000) * 1000, (p / 100) * 100, r);
        } else
          printf("4\n%d %d %d %d\n", m * 1000, p - r, r - (r % 10), n % 10);
      }
    }
  }
}