#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 3;
inline int readint() {
  int ans = 0;
  bool sign = true;
  char c;
  while (!isdigit(c = getchar()))
    if (c == '-') sign = false;
  while (isdigit(c)) {
    ans = (ans << 3) + (ans << 1) + (c ^ 48);
    c = getchar();
  }
  return ans * (sign ? 1 : -1);
}
inline void readint(int &x) { x = readint(); }
template <typename... Ts>
inline void readint(int &x, Ts &...y) {
  x = readint();
  readint(y...);
}
const int INF = 0x3f3f3f3f;
const long long LLINF = ((long long)INF << 32) | INF;
const int mod = 1e9 + 7;
int main() {
  ;
  set<int> pow10;
  for (int i = 1; i <= 1e9; i *= 10) pow10.insert(i);
  int ___;
  cin >> ___;
  for (int tc = 1; tc <= ___; ++tc) {
    int s, n;
    readint(s, n);
    vector<int> ans;
    ans.push_back(s);
    while (ans.size() < n) {
      int no10 = -1;
      for (int i = 0; i < ans.size(); ++i) {
        if (!pow10.count(ans[i])) {
          no10 = i;
          break;
        }
      }
      if (no10 != -1) {
        int t = 1e9;
        while (ans[no10] <= t) {
          t /= 10;
        }
        ans.push_back(t);
        ans[no10] -= t;
      } else {
        sort(begin(ans), end(ans));
        int mx = -1;
        for (int i = 0; i < ans.size(); ++i) {
          if (ans[i] != 1) {
            mx = i;
            break;
          }
        }
        int t = 1e9;
        while (ans[mx] <= t) {
          t /= 10;
        }
        ans.push_back(t);
        ans[mx] -= t;
      }
    }
    for (int i : ans) printf("%d ", i);
    puts("");
  }
  return 0;
}

### Prompt

Generate 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 int N = 2e5 + 3;
inline int readint() {
  int ans = 0;
  bool sign = true;
  char c;
  while (!isdigit(c = getchar()))
    if (c == '-') sign = false;
  while (isdigit(c)) {
    ans = (ans << 3) + (ans << 1) + (c ^ 48);
    c = getchar();
  }
  return ans * (sign ? 1 : -1);
}
inline void readint(int &x) { x = readint(); }
template <typename... Ts>
inline void readint(int &x, Ts &...y) {
  x = readint();
  readint(y...);
}
const int INF = 0x3f3f3f3f;
const long long LLINF = ((long long)INF << 32) | INF;
const int mod = 1e9 + 7;
int main() {
  ;
  set<int> pow10;
  for (int i = 1; i <= 1e9; i *= 10) pow10.insert(i);
  int ___;
  cin >> ___;
  for (int tc = 1; tc <= ___; ++tc) {
    int s, n;
    readint(s, n);
    vector<int> ans;
    ans.push_back(s);
    while (ans.size() < n) {
      int no10 = -1;
      for (int i = 0; i < ans.size(); ++i) {
        if (!pow10.count(ans[i])) {
          no10 = i;
          break;
        }
      }
      if (no10 != -1) {
        int t = 1e9;
        while (ans[no10] <= t) {
          t /= 10;
        }
        ans.push_back(t);
        ans[no10] -= t;
      } else {
        sort(begin(ans), end(ans));
        int mx = -1;
        for (int i = 0; i < ans.size(); ++i) {
          if (ans[i] != 1) {
            mx = i;
            break;
          }
        }
        int t = 1e9;
        while (ans[mx] <= t) {
          t /= 10;
        }
        ans.push_back(t);
        ans[mx] -= t;
      }
    }
    for (int i : ans) printf("%d ", i);
    puts("");
  }
  return 0;
}

```