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;
  void in(int &n) {
    n = 0;
    int f = 1;
    char c = getchar();
    while (c < '0' || '9' < c) {
      if (c == '-') f *= -1;
      c = getchar();
    }
    while ('0' <= c && c <= '9') {
      n = n * 10 + (c - '0');
      c = getchar();
    }
    n *= f;
  }
  void llin(long long &n) {
    n = 0;
    long long f = 1;
    char c = getchar();
    while (c < '0' || '9' < c) {
      if (c == '-') f *= -1;
      c = getchar();
    }
    while ('0' <= c && c <= '9') {
      n = n * 10 + (c - '0');
      c = getchar();
    }
    n *= f;
  }
  long long pw10[25], a[25], b[25], sum[25];
  int main() {
    pw10[0] = 1;
    for (int i = 1; i <= 18; i++) pw10[i] = pw10[i - 1] * 10ll;
    int t;
    in(t);
    while (t--) {
      memset(a, 0, sizeof(a));
      memset(b, 0, sizeof(b));
      memset(sum, 0, sizeof(sum));
      int n;
      long long m;
      in(n);
      llin(m);
      m++;
      for (int i = 1; i <= n; i++) {
        int x;
        in(x);
        a[x] = 1;
        for (int j = x + 1; j <= 18; j++) a[j] = a[j - 1] * 10ll;
      }
      sum[0] = a[0] * 9ll;
      for (int i = 1; i <= 18; i++) sum[i] = sum[i - 1] + 9ll * a[i];
      while (m) {
        if (m <= sum[0]) {
          long long tmp = m;
          tmp = tmp / a[0] + ((tmp % a[0]) != 0);
          b[0] = tmp;
          m -= tmp * a[0];
          continue;
        }
        for (int i = 0; i < 18; i++) {
          if (sum[i] <= m && m <= sum[i + 1]) {
            long long tmp = m - sum[i];
            tmp = tmp / a[i + 1] + ((tmp % a[i + 1]) != 0);
            b[i + 1] = tmp;
            m -= tmp * a[i + 1];
            break;
          }
        }
      }
      long long ans = 0;
      for (int i = 18; i >= 0; i--) ans = ans * 10ll + b[i];
      printf("%lld\n", ans);
    }
    return 0;
  }
  