description: 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 $$$3_{11} + 4_{11} + 100_{11} + 4_{11} = 110_{11} = 1 ⋅ 11^2 + 1 ⋅ 11 = 132_{10}. It can be shown that it is impossible for Alice to get a larger sum than 132_{10}$$$.
solution: #include <bits/stdc++.h>
  using namespace std;
  long long pow2(long long i) { return 1LL << i; }
  long long topbit(signed t) { return t == 0 ? -1 : 31 - __builtin_clz(t); }
  long long topbit(long long t) { return t == 0 ? -1 : 63 - __builtin_clzll(t); }
  long long lowbit(signed a) { return a == 0 ? 32 : __builtin_ctz(a); }
  long long lowbit(long long a) { return a == 0 ? 64 : __builtin_ctzll(a); }
  long long allbit(long long n) { return (1LL << n) - 1; }
  long long popcount(signed t) { return __builtin_popcount(t); }
  long long popcount(long long t) { return __builtin_popcountll(t); }
  bool ispow2(long long i) { return i && (i & -i) == i; }
  template <class T>
  T POW(T x, long long n) {
    T res = 1;
    for (; n; n >>= 1, x *= x)
      if (n & 1) res *= x;
    return res;
  }
  inline long long max(long long x, long long y) { return (x < y) ? y : x; }
  inline long long min(long long x, long long y) { return (x < y) ? x : y; }
  inline long long read() {
    register long long a = 0, po = 1;
    char ch = getchar();
    while (!isdigit(ch) && ch != '-') ch = getchar();
    if (ch == '-') po = -1, ch = getchar();
    while (isdigit(ch)) a = (a << 1) + (a << 3) + ch - 48, ch = getchar();
    return a * po;
  }
  void write(long long x) {
    if (x < 0) putchar('-'), x = -x;
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
    return;
  }
  long long gcd(long long x, long long y) {
    while (x % y) {
      long long temp = x % y;
      x = y, y = temp;
    }
    return y;
  }
  long long mgml(long long a, long long b, long long mod) {
    long long temp = 1;
    while (b) {
      if (b % 2) {
        temp = temp * a % mod;
      }
      a = a * a % mod;
      b /= 2;
    }
    return temp;
  }
  long long exgcd(long long a, long long b, long long &x, long long &y) {
    long long ret, tmp;
    if (!b) {
      x = 1;
      y = 0;
      return a;
    }
    ret = exgcd(b, a % b, x, y);
    tmp = x;
    x = y;
    y = tmp - a / b * y;
    return ret;
  }
  long long lcm(long long x, long long y) { return (x * y) / gcd(x, y); }
  long long fac[100009], inv[100009], MXNC = 100005;
  long long mod = 1000000007;
  void init() {
    inv[0] = inv[1] = fac[0] = 1;
    for (long long i = 2; i <= MXNC; i++)
      inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
    for (long long i = 2; i <= MXNC; i++)
      inv[i] = 1ll * inv[i - 1] * inv[i] % mod;
    for (long long i = 1; i < MXNC; i++) fac[i] = fac[i - 1] * i % mod;
  }
  long long C(long long n, long long m) {
    return 1ll * fac[n] * inv[m] % mod * inv[n - m] % mod;
  }
  long long fenkuai(long long n) {
    long long ans = 0;
    for (long long l = 1, r; l <= n; l = r + 1) {
      r = n / (n / l);
      ans += (r - l + 1) * (n / l);
    }
    return ans;
  }
  long long fl[1200010];
  vector<long long> prime;
  void gep() {
    memset(fl, false, sizeof(fl));
    fl[1] = true;
    for (long long i = 2; i <= 1200000; i++) {
      long long t = 1200000 / i;
      for (long long j = 2; j <= t; j++) fl[i * j] = true;
    }
    for (long long i = 2; i <= 1200000; i++)
      if (!fl[i]) prime.push_back(i);
  }
  signed main() {
    long long t = read();
    while (t--) {
      long long n = read(), m = read();
      for (long long i = 1; i <= m; i++) {
        if (i == m)
          cout << n << "\n";
        else if (n > 1000000000 && n - 1000000000 >= m - i) {
          cout << 1000000000 << " ";
          n -= 1000000000;
        } else if (n > 100000000 && n - 100000000 >= m - i) {
          cout << 100000000 << " ";
          n -= 100000000;
        } else if (n > 10000000 && n - 10000000 >= m - i) {
          cout << 10000000 << " ";
          n -= 10000000;
        } else if (n > 1000000 && n - 1000000 >= m - i) {
          cout << 1000000 << " ";
          n -= 1000000;
        } else if (n > 100000 && n - 100000 >= m - i) {
          cout << 100000 << " ";
          n -= 100000;
        } else if (n > 10000 && n - 10000 >= m - i) {
          cout << 10000 << " ";
          n -= 10000;
        } else if (n > 1000 && n - 1000 >= m - i) {
          cout << 1000 << " ";
          n -= 1000;
        } else if (n > 100 && n - 100 >= m - i) {
          cout << 100 << " ";
          n -= 100;
        } else if (n > 10 && n - 10 >= m - i) {
          cout << 10 << " ";
          n -= 10;
        } else {
          cout << 1 << " ";
          n--;
        }
      }
    }
    return 0;
  }
  