#include <bits/stdc++.h>
using namespace std;
void Fast() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
void File() {}
using ll = long long;
using ull = unsigned long long;
int dx[] = {0, 0, -1, 1, 1, 1, -1, -1};
int dy[] = {1, -1, 0, 0, 1, -1, 1, -1};
const int N = 1e6 + 5;
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> v(n);
  for (auto& a : v) cin >> a;
  string ans;
  int bit = 0;
  ll num = 9;
  k++;
  for (int i = 0; k && i < n - 1; i++, num = 9, bit++) {
    while (v[i + 1] != bit + 1) {
      num = num * 10 + 9;
      bit++;
    }
    if (i == n) break;
    if (k <= num) {
      ans = to_string(k) + ans;
      k = 0;
      break;
    } else {
      ans = to_string(num) + ans;
      k -= num;
    }
  }
  if (k) cout << k;
  cout << ans << "\n";
}
int main() {
  Fast();
  File();
  int t$$t$;
  cin >> t$$t$;
  while (t$$t$--) solve();
}

### Prompt

Please formulate 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 Fast() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
void File() {}
using ll = long long;
using ull = unsigned long long;
int dx[] = {0, 0, -1, 1, 1, 1, -1, -1};
int dy[] = {1, -1, 0, 0, 1, -1, 1, -1};
const int N = 1e6 + 5;
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> v(n);
  for (auto& a : v) cin >> a;
  string ans;
  int bit = 0;
  ll num = 9;
  k++;
  for (int i = 0; k && i < n - 1; i++, num = 9, bit++) {
    while (v[i + 1] != bit + 1) {
      num = num * 10 + 9;
      bit++;
    }
    if (i == n) break;
    if (k <= num) {
      ans = to_string(k) + ans;
      k = 0;
      break;
    } else {
      ans = to_string(num) + ans;
      k -= num;
    }
  }
  if (k) cout << k;
  cout << ans << "\n";
}
int main() {
  Fast();
  File();
  int t$$t$;
  cin >> t$$t$;
  while (t$$t$--) solve();
}

```