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 MOD = 1e18 + 100;
const int maxn = 1e4 + 100;
const int len = 35;
const long long linf = 9223372036854775807;
const int inf = 0x3f3f3f3f;
long long qpow(long long a, long long b) {
  return b ? ((b & 1) ? a * qpow(a * a % MOD, b >> 1) % MOD
                      : qpow(a * a % MOD, b >> 1)) %
                 MOD
           : 1;
}
long long qpow(long long a, long long b, long long c) {
  return b ? ((b & 1) ? a * qpow(a * a % MOD, b >> 1) % c
                      : qpow(a * a % MOD, b >> 1)) %
                 c
           : 1;
}
inline int ird() {
  register int x = 0, f = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return x * f;
}
inline long long lrd() {
  register long long x = 0, f = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return x * f;
}
void puts1() { cout << "-1" << endl; }
void puts2() { cout << "YES" << endl; }
void puts3() { cout << "NO" << endl; }
int gcd(int a, int b) { return (b > 0) ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a * b / gcd(a, b); }
long long lgcd(long long a, long long b) {
  return (b > 0) ? lgcd(b, a % b) : a;
}
long long llcm(long long a, long long b) { return a * b / lgcd(a, b); }
int f[maxn * 3];
int fin(int x) {
  if (f[x] == x) return x;
  return f[x] = fin(f[x]);
}
void mix(int x, int y) { f[fin(x)] = fin(y); }
int a[maxn];
string s;
int main() {
  int n, m, ju, b, fa, fb;
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    int co = 0, tt = 1;
    while (n) {
      if (n % 10 == 0) {
        tt *= 10;
        n /= 10;
        continue;
      }
      a[++co] = (n % 10) * tt;
      n /= 10;
      tt *= 10;
    }
    cout << co << endl;
    for (int i = co; i >= 1; i--) cout << a[i] << " ";
    cout << endl;
  }
}