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;
  const int N = 15;
  int a[N];
  long long ksm(long long a, long long b, long long c) {
    int res = 1;
    while (b) {
      if (b & 1) res = (res * a) % c;
      a = (a * a) % c;
      b >>= 1;
    }
    return res % c;
  }
  int read() {
    int n;
    cin >> n;
    return n;
  }
  void solve() {
    int n;
    long long k;
    cin >> n >> k;
    for (int i = 1; i <= n; i++) cin >> a[i];
    int pos = 0;
    vector<long long> v;
    bool f = 0;
    for (int i = 1; i <= n; i++) {
      if (i == 1 && a[i] != 0) {
        cout << 0 << '\n';
        return;
      }
      v.push_back(min(ksm(10, a[i] - pos, 1e18) - 1, k));
      if (k + 1 == ksm(10, a[i] - pos, 1e18)) f = true;
      k = max(0ll, k - ksm(10, a[i] - pos, 1e18) + 1);
      if (k == 0) break;
      pos = a[i];
    }
    if (v.size() == 1) {
      cout << v[0] + k + 1 << '\n';
      return;
    }
    if (k != 0) {
      v.push_back(k);
      v.back()++;
      for (int i = v.size() - 1; i > 0; i--) cout << v[i];
      cout << '\n';
      return;
    }
    if (k == 0) {
      if (!f) {
        v.back()++;
      } else
        v.push_back(1);
    }
    for (int i = v.size() - 1; i > 0; i--) cout << v[i];
    cout << '\n';
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int _;
    cin >> _;
    while (_--) {
      solve();
    }
    return 0;
  }
  