#include <bits/stdc++.h>
using namespace std;
void setIO() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
const int64_t N = 1e5 + 5;
const int64_t M = 650;
const int64_t inf = 9e12;
const int64_t mod = 998244353;
int64_t xyz = 1;
int64_t n, m;
int64_t arr[N];
int64_t pwr[N];
void run() {
  cin >> n >> m;
  m++;
  pwr[0] = 1;
  for (int64_t i = 1; i <= 12; i++) {
    pwr[i] = pwr[i - 1] * 10;
  }
  for (int64_t i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  arr[n + 1] = 12;
  int64_t num = 0;
  int64_t sum = 0;
  for (int64_t i = 1; i <= n; i++) {
    int64_t add = pwr[arr[i + 1] - arr[i]] - 1;
    if (i == n || sum + add > m) {
      num += (m - sum) * pwr[arr[i]];
      break;
    } else
      sum += add, num += pwr[arr[i]] * add;
  }
  cout << num << "\n";
}
signed main() {
  setIO();
  cin >> xyz;
  while (xyz--) run();
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 setIO() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
const int64_t N = 1e5 + 5;
const int64_t M = 650;
const int64_t inf = 9e12;
const int64_t mod = 998244353;
int64_t xyz = 1;
int64_t n, m;
int64_t arr[N];
int64_t pwr[N];
void run() {
  cin >> n >> m;
  m++;
  pwr[0] = 1;
  for (int64_t i = 1; i <= 12; i++) {
    pwr[i] = pwr[i - 1] * 10;
  }
  for (int64_t i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  arr[n + 1] = 12;
  int64_t num = 0;
  int64_t sum = 0;
  for (int64_t i = 1; i <= n; i++) {
    int64_t add = pwr[arr[i + 1] - arr[i]] - 1;
    if (i == n || sum + add > m) {
      num += (m - sum) * pwr[arr[i]];
      break;
    } else
      sum += add, num += pwr[arr[i]] * add;
  }
  cout << num << "\n";
}
signed main() {
  setIO();
  cin >> xyz;
  while (xyz--) run();
  return 0;
}

```