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 long long mod = 1e9 + 7;
  int T;
  long long n, k;
  long long cost[20], b[20], a[20];
  int main() {
    scanf("%d", &T);
    b[0] = 1;
    for (int i = 1; i <= 10; ++i) {
      b[i] = b[i - 1] * 10;
    }
    while (T--) {
      scanf("%lld%lld", &n, &k);
      k++;
      memset(cost, 0, sizeof(cost));
      memset(a, 0, sizeof(a));
      cost[0] = 1;
      for (int i = 1; i <= n; ++i) {
        int bits;
        scanf("%d", &bits);
        cost[bits] = 1;
      }
      for (int i = 1; i <= 10; ++i) {
        if (!cost[i]) {
          cost[i] = cost[i - 1] * 10;
        }
      }
      long long ans = 0, tot = 0;
      for (int i = 0; i <= 9; ++i) {
        k -= 9 * cost[i];
        a[i] = 9;
        ans += 9 * b[i];
        if (k <= 0) break;
      }
      if (k > 0) {
        ans += (k / cost[10]) * b[10];
        a[10] += (k / cost[10]);
        if (k % cost[10]) {
          ans += b[10];
          a[10]++;
        }
        k -= cost[10] * a[10];
      }
      for (int i = 10; i >= 0; --i) {
        while (a[i] != 0 && (-k) >= cost[i]) {
          ans -= b[i];
          a[i]--;
          k += cost[i];
        }
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  