description: In Berland, n different types of banknotes are used. Banknotes of the i-th type have denomination 10^{a_i} burles (burles are the currency used in Berland); the denomination of banknotes of the first type is exactly 1.
  
  Let's denote f(s) as the minimum number of banknotes required to represent exactly s burles. For example, if the denominations of banknotes used in Berland are 1, 10 and 100, then f(59) = 14: 9 banknotes with denomination of 1 burle and 5 banknotes with denomination of 10 burles can be used to represent exactly 9 ⋅ 1 + 5 ⋅ 10 = 59 burles, and there's no way to do it with fewer banknotes.
  
  For a given integer k, find the minimum positive number of burles s that cannot be represented with k or fewer banknotes (that is, f(s) > k).
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
  
  The first line of each test case contains two integers n and k (1 ≤ n ≤ 10; 1 ≤ k ≤ 10^9).
  
  The next line contains n integers a_1, a_2, ..., a_n (0 = a_1 < a_2 < ... < a_n ≤ 9).
  
  Output
  
  For each test case, print one integer — the minimum positive number of burles s that cannot be represented with k or fewer banknotes.
  
  Example
  
  Input
  
  
  4
  3 13
  0 1 2
  2 777
  0 4
  3 255
  0 1 3
  10 1000000000
  0 1 2 3 4 5 6 7 8 9
  
  
  Output
  
  
  59
  778
  148999
  999999920999999999
solution: #include <bits/stdc++.h>
  using namespace std;
  template <typename T, typename U>
  istream& operator>>(istream& in, pair<T, U>& a) {
    in >> a.first >> a.second;
    return in;
  }
  template <typename T, typename U>
  ostream& operator<<(ostream& out, pair<T, U> a) {
    out << '(' << a.first << ", " << a.second << ')';
    return out;
  }
  template <typename T>
  ostream& operator<<(ostream& cout, vector<T> const& v) {
    cout << "[";
    for (long long i = 0; i < (long long)((v).size()); i++) {
      if (i) cout << ", ";
      cout << v[i];
    }
    return cout << "]";
  }
  template <typename T, typename U>
  static inline void amin(T& x, U y) {
    if (y < x) x = y;
  }
  template <typename T, typename U>
  static inline void amax(T& x, U y) {
    if (x < y) x = y;
  }
  const long double pi = 3.14159265358979323846;
  const char nl = '\n';
  const long long INF = 1e18;
  const int32_t M = 1e9 + 7;
  const int32_t MM = 998244353;
  const long long N = 1e6 + 5;
  long long n, m, q, k, l, r, x, y, z, a[N], b[N], c[N];
  string second, t;
  void solve() {
    cin >> n >> k;
    vector<long long> pw(20);
    pw[0] = 1;
    for (long long i = 1; i < 20; i++) pw[i] = pw[i - 1] * 10;
    vector<long long> a(20), r(20);
    for (long long i = 0; i < n; i++) {
      cin >> x;
      a[x] = 1;
    }
    vector<long long> c(20);
    for (long long i = 0; i < 20; i++) {
      if (i) r[i] = r[i - 1];
      if (a[i]) r[i] = i;
      c[i] = pw[i - r[i]];
      if (i) c[i] += c[i - 1];
    }
    for (long long i = 0; i < 20; i++) {
      if (9 * c[i] <= k) continue;
      long long num = 0, cost = 0;
      for (long long j = i; j >= 0; j--) {
        for (long long h = 0; h < 10; h++) {
          if (cost + h * pw[j - r[j]] + (j ? c[j - 1] * 9 : 0) > k) {
            num += h * pw[j];
            cost += h * pw[j - r[j]];
            break;
          }
        }
      }
      cout << num << nl;
      return;
    }
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t = 1;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  