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;
  using ll = int64_t;
  using ull = uint64_t;
  using ld = long double;
  using vi = vector<int64_t>;
  using pii = pair<int64_t, int64_t>;
  ll powr(ll a, ll b) {
    ll res = 1;
    while (b) {
      if (b & 1) res = (res * a);
      a = (a * a);
      b >>= 1;
    }
    return res;
  }
  void solve() {
    int64_t n, k;
    cin >> n >> k;
    k++;
    int64_t a[n];
    for (int64_t i = 0; i < n; i++) {
      cin >> a[i];
      a[i] = powr(10, a[i]);
    }
    sort(a, a + n);
    int64_t ans = 0;
    for (int64_t i = 0; i < n - 1 && k > 0; i++) {
      int64_t den = min(k, (a[i + 1] / a[i]) - 1);
      ans += a[i] * den;
      k -= den;
    }
    if (k > 0) {
      ans += a[n - 1] * k;
    }
    cout << ans << '\n';
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int64_t t;
    cin >> t;
    while (t--) solve();
  }
  