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 ll = long long;
  using namespace std;
  const int maxn = 1e3 + 5;
  ll ksm(ll a, ll n) {
    ll ans = 1;
    while (n) {
      if (n % 2) ans = (a * ans);
      a = (a * a);
      n /= 2;
    }
    return ans;
  }
  void solve() {
    ll a[maxn], dp[maxn], ten[maxn], n, k, now;
    for (int i = 0; i <= 10; i++) {
      ten[i] = ksm(10, i) - 1;
    }
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    now = n + 1;
    for (int i = 2; i <= n; i++) {
      dp[i] = ten[a[i] - a[i - 1]] + dp[i - 1];
      if (k < dp[i]) {
        now = i;
        break;
      }
    }
    cout << (ten[a[now - 1]] + 1) * (k - dp[now - 1] + 2) - 1 << endl;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  