#include <bits/stdc++.h>
using namespace std;
void _out() { cerr << '\n'; }
template <typename H, typename... T>
void _out(H h, T... t) {
  cerr << ' ' << h;
  _out(t...);
}
void solve();
void init();
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(0);
  long long T = 1;
  cin >> T;
  init();
  while (T--) {
    solve();
  }
  return 0;
}
long long const N = 20;
long long a[N];
long long ten[N];
void init() {
  long long t = 1;
  for (long long i = 0; i < N; i++) {
    ten[i] = t;
    t *= 10;
  }
}
void solve() {
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long ans = 0;
  k++;
  for (long long i = 0; i < n; i++) {
    if (i != n - 1 && k >= ten[a[i + 1] - a[i]] - 1) {
      ans += (ten[a[i + 1] - a[i]] - 1) * ten[a[i]];
      k -= ten[a[i + 1] - a[i]] - 1;
    } else {
      ans += k * ten[a[i]];
      break;
    }
  }
  cout << ans << '\n';
}

### Prompt

Generate a Cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void _out() { cerr << '\n'; }
template <typename H, typename... T>
void _out(H h, T... t) {
  cerr << ' ' << h;
  _out(t...);
}
void solve();
void init();
signed main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(0);
  long long T = 1;
  cin >> T;
  init();
  while (T--) {
    solve();
  }
  return 0;
}
long long const N = 20;
long long a[N];
long long ten[N];
void init() {
  long long t = 1;
  for (long long i = 0; i < N; i++) {
    ten[i] = t;
    t *= 10;
  }
}
void solve() {
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long ans = 0;
  k++;
  for (long long i = 0; i < n; i++) {
    if (i != n - 1 && k >= ten[a[i + 1] - a[i]] - 1) {
      ans += (ten[a[i + 1] - a[i]] - 1) * ten[a[i]];
      k -= ten[a[i + 1] - a[i]] - 1;
    } else {
      ans += k * ten[a[i]];
      break;
    }
  }
  cout << ans << '\n';
}

```