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;
  long long maxn = 2000001;
  long long mod = 1000000007;
  vector<long long> sieveVector(maxn + 1, 1);
  vector<long long> factVector(maxn + 1, 1);
  long long power(long long a, long long b) {
    long long res = 1;
    while (b) {
      if (b % 2)
        res = res * a, b--;
      else
        a = a * a, b /= 2;
    }
    return res;
  }
  long long powerMod(long long a, long long b, long long m) {
    long long res = 1;
    while (b) {
      if (b % 2)
        res = (res % m * a % m) % m, b--;
      else
        a = (a % m * a % m) % m, b /= 2;
    }
    return res;
  }
  long long modIn(long long n, long long m) { return powerMod(n, m - 2, m); }
  void factorial() {
    for (long long i = 1; i <= maxn; i++)
      factVector[i] = (i % mod * factVector[i - 1] % mod);
  }
  void sieve() {
    for (long long i = 2; i * i <= maxn; i++) {
      if (sieveVector[i]) {
        for (long long j = i * i; j <= maxn; j += i) {
          sieveVector[j] = 0;
        }
      }
    }
  }
  long long ncr(long long n, long long r) {
    if (r > n) return 0;
    return (factVector[n] % mod *
            modIn((factVector[n - r] % mod * factVector[r] % mod) % mod, mod)) %
           mod;
  }
  void solve() {
    long long n, k;
    cin >> n >> k;
    vector<long long> v(n);
    for (long long I = 0; I < n; I++) cin >> v[I];
    ;
    long long ans = 0;
    k++;
    for (long long i = 0; i < n - 1; i++) {
      if (!k) break;
      long long d = power(10, v[i + 1]) - 1;
      long long maxe = d / power(10, v[i]);
      ans += min(k, maxe) * power(10, v[i]);
      k -= min(k, maxe);
    }
    ans += (k * power(10, v.back()));
    cout << ans << endl;
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long tc;
    cin >> tc;
    while (tc--) solve();
  }
  