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 maxn = 100;
  long long n, k;
  int a[15];
  long long rc[15], p[15], ans;
  void solve() {
    scanf("%d%d", &n, &k);
    k++;
    scanf("%d", &a[1]);
    rc[1] = 1;
    long long now = 1, tmp;
    int ss = 0;
    for (int i = 2; i <= n; i++) {
      scanf("%d", &a[i]);
      tmp = 1;
      while (ss < a[i]) now *= 10, tmp *= 10, ss++;
      p[i - 1] = tmp - 1;
      rc[i] = now;
    }
    ans = 0;
    for (int i = 1; i < n; i++) {
      if (k >= p[i])
        ans += p[i] * rc[i], k -= p[i];
      else {
        ans += rc[i] * k;
        k = 0;
      }
    }
    if (k) ans += rc[n] * k;
    printf("%lld\n", ans);
  }
  int main() {
    int t;
    scanf("%d", &t);
    while (t--) solve();
    return 0;
  }
  