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;
  int a[10];
  long long v[10];
  long long pw[12];
  int len(long long x) {
    int ans = 0;
    while (x) {
      ans++;
      x /= 10;
    }
    return ans;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    pw[0] = 1;
    for (int i = 1; i <= 10; i++) pw[i] = pw[i - 1] * 10;
    while (t--) {
      long long n, k;
      cin >> n >> k;
      k++;
      for (int i = 0; i < n; i++) cin >> a[i];
      for (int i = 1; i < n; i++) {
        v[i] = min(k, pw[a[i] - a[i - 1]] - 1);
        k -= v[i];
      }
      bool flag = false;
      if (k != 0) {
        cout << k;
        flag = true;
      }
      for (int i = n - 2; i >= 0; i--) {
        if (flag)
          for (int j = 1; j <= a[i + 1] - a[i] - len(v[i + 1]); j++) cout << "0";
        if (v[i + 1] != 0) {
          cout << v[i + 1];
          flag = true;
        }
      }
      cout << "\n";
    }
  }
  