#include <bits/stdc++.h>
using namespace std;
const long long large = 1e10;
const long long remi = 1000000007;
const long long remi2 = 998244353;
const long long inf = 1e18 + 1e17 + 1e16 + 1e15 + 1e14;
vector<vector<int> > g;
vector<int> vis;
long long PowI(long long a, long long b, long long m) {
  long long ans = 1 % m;
  while (b > 0) {
    if (b % 2) ans = (((ans % m) * (a % m)) % m);
    a = (((a % m) * (a % m)) % m);
    b = (long long)(b / ((long long)2));
  }
  return ans;
}
vector<long long> sepdig(long long n) {
  vector<long long> ans;
  while (n) {
    long long r = n % 10;
    n /= 10;
    ans.push_back(r);
  }
  reverse(ans.begin(), ans.end());
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long i, j, k;
  int T = 1;
  cin >> T;
  while (T--) {
    long long n, ele;
    cin >> n >> ele;
    vector<long long> digs = sepdig(n);
    long long dig = digs.size();
    multiset<long long> spes;
    for (i = 0; i < digs.size(); i++) {
      long long num = digs[i];
      long long num2 = PowI(10, dig - 1 - i, inf);
      for (j = 0; j < num; j++) {
        spes.insert(num2);
      }
    }
    long long sz = spes.size();
    vector<long long> ans;
    if (sz < ele) {
      while (sz < ele) {
        for (auto x = spes.begin(); x != spes.end(); x++) {
          long long num = *x;
          if (num != 1) {
            long long numins = num / 10;
            spes.erase(spes.find(num));
            for (j = 0; j < 10; j++) spes.insert(numins);
            break;
          }
        }
        sz = spes.size();
      }
    }
    long long sum = n;
    i = 0;
    for (auto x = spes.begin(); x != spes.end(); x++) {
      if (i >= ele - 1) break;
      long long num = *x;
      ans.push_back(num);
      sum -= num;
      i++;
      if (i >= ele - 1) break;
    }
    ans.push_back(sum);
    sort(ans.begin(), ans.end(), greater<long long>());
    for (i = 0; i < ele; i++) cout << ans[i] << " ";
    cout << "\n";
  }
}

### Prompt

Please formulate a Cpp solution to the following problem:

On the board, Bob wrote n positive integers in [base](https://en.wikipedia.org/wiki/Positional_notation#Base_of_the_numeral_system) 10 with sum s (i. e. in decimal numeral system). Alice sees the board, but accidentally interprets the numbers on the board as base-11 integers and adds them up (in base 11).

What numbers should Bob write on the board, so Alice's sum is as large as possible?

Input

The input consists of multiple test cases. The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.

The only line of each test case contains two integers s and n (1 ≤ s ≤ 10^9; 1 ≤ n ≤ min(100, s)) — the sum and amount of numbers on the board, respectively. Numbers s and n are given in decimal notation (base 10).

Output

For each test case, output n positive integers — the numbers Bob should write on the board, so Alice's sum is as large as possible. If there are multiple answers, print any of them.

Example

Input


6
97 2
17 1
111 4
100 2
10 9
999999 3


Output


70 27 
17 
3 4 100 4
10 90
1 1 2 1 1 1 1 1 1 
999900 90 9

Note

In the first test case, 70_{10} + 27_{10} = 97_{10}, and Alice's sum is $$$70_{11} + 27_{11} = 97_{11} = 9 ⋅ 11 + 7 = 106_{10}. (Here x_b represents the number x in base b.) It can be shown that it is impossible for Alice to get a larger sum than 106_{10}$$$.

In the second test case, Bob can only write a single number on the board, so he must write 17.

In the third test case, 3_{10} + 4_{10} + 100_{10} + 4_{10} = 111_{10}, and Alice's sum is $$$3_{11} + 4_{11} + 100_{11} + 4_{11} = 110_{11} = 1 ⋅ 11^2 + 1 ⋅ 11 = 132_{10}. It can be shown that it is impossible for Alice to get a larger sum than 132_{10}$$$.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long large = 1e10;
const long long remi = 1000000007;
const long long remi2 = 998244353;
const long long inf = 1e18 + 1e17 + 1e16 + 1e15 + 1e14;
vector<vector<int> > g;
vector<int> vis;
long long PowI(long long a, long long b, long long m) {
  long long ans = 1 % m;
  while (b > 0) {
    if (b % 2) ans = (((ans % m) * (a % m)) % m);
    a = (((a % m) * (a % m)) % m);
    b = (long long)(b / ((long long)2));
  }
  return ans;
}
vector<long long> sepdig(long long n) {
  vector<long long> ans;
  while (n) {
    long long r = n % 10;
    n /= 10;
    ans.push_back(r);
  }
  reverse(ans.begin(), ans.end());
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long i, j, k;
  int T = 1;
  cin >> T;
  while (T--) {
    long long n, ele;
    cin >> n >> ele;
    vector<long long> digs = sepdig(n);
    long long dig = digs.size();
    multiset<long long> spes;
    for (i = 0; i < digs.size(); i++) {
      long long num = digs[i];
      long long num2 = PowI(10, dig - 1 - i, inf);
      for (j = 0; j < num; j++) {
        spes.insert(num2);
      }
    }
    long long sz = spes.size();
    vector<long long> ans;
    if (sz < ele) {
      while (sz < ele) {
        for (auto x = spes.begin(); x != spes.end(); x++) {
          long long num = *x;
          if (num != 1) {
            long long numins = num / 10;
            spes.erase(spes.find(num));
            for (j = 0; j < 10; j++) spes.insert(numins);
            break;
          }
        }
        sz = spes.size();
      }
    }
    long long sum = n;
    i = 0;
    for (auto x = spes.begin(); x != spes.end(); x++) {
      if (i >= ele - 1) break;
      long long num = *x;
      ans.push_back(num);
      sum -= num;
      i++;
      if (i >= ele - 1) break;
    }
    ans.push_back(sum);
    sort(ans.begin(), ans.end(), greater<long long>());
    for (i = 0; i < ele; i++) cout << ans[i] << " ";
    cout << "\n";
  }
}

```