#include <bits/stdc++.h>
using namespace std;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
const int maxN = 1e5 + 10;
int n, a[maxN];
long long k, pw[19];
void solve() {
  cin >> n >> k;
  for (int i = 0; i < (int)(n); ++i) {
    cin >> a[i];
  }
  pw[0] = 1;
  for (int i = 1; i <= (int)(18); ++i) {
    pw[i] = pw[i - 1] * 10;
  }
  k++;
  string ans;
  for (int i = 1; i <= (int)(n - 1); ++i) {
    int d = a[i] - a[i - 1];
    if (pw[d] - 1 < k) {
      ans += string(d, '9');
      k -= pw[d] - 1;
    } else {
      string t = to_string(k);
      reverse((t).begin(), (t).end());
      ans += t;
      k = 0;
      break;
    }
  }
  if (k) {
    string t = to_string(k);
    reverse((t).begin(), (t).end());
    ans += t;
  }
  reverse((ans).begin(), (ans).end());
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
template <class T>
bool uin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
const int maxN = 1e5 + 10;
int n, a[maxN];
long long k, pw[19];
void solve() {
  cin >> n >> k;
  for (int i = 0; i < (int)(n); ++i) {
    cin >> a[i];
  }
  pw[0] = 1;
  for (int i = 1; i <= (int)(18); ++i) {
    pw[i] = pw[i - 1] * 10;
  }
  k++;
  string ans;
  for (int i = 1; i <= (int)(n - 1); ++i) {
    int d = a[i] - a[i - 1];
    if (pw[d] - 1 < k) {
      ans += string(d, '9');
      k -= pw[d] - 1;
    } else {
      string t = to_string(k);
      reverse((t).begin(), (t).end());
      ans += t;
      k = 0;
      break;
    }
  }
  if (k) {
    string t = to_string(k);
    reverse((t).begin(), (t).end());
    ans += t;
  }
  reverse((ans).begin(), (ans).end());
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

```