#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (x < y) x = y;
}
template <typename... T>
void read(T &...args) {
  ((cin >> args), ...);
}
template <typename... T>
void write(T &&...args) {
  ((cout << args << endl), ...);
}
const long long int mod = 1e9 + 7, N = 1e5 + 1;
vector<pair<long long int, long long int>> moves = {
    {-1, 0}, {1, 0}, {0, -1}, {0, 1}};
void solve() {
  long long int n, k;
  cin >> n >> k;
  k++;
  long long int a[n];
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long int ans = 0;
  for (long long int i = 0; i < n - 1; i++) {
    long long int noe = pow(10, a[i + 1] - a[i]) - 1;
    if (k > noe) {
      ans += noe * pow(10, a[i]);
      k -= noe;
    } else {
      ans += k * pow(10, a[i]);
      k = 0;
      break;
    }
  }
  ans += k * pow(10, a[n - 1]);
  cout << ans << endl;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 <typename T, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (x < y) x = y;
}
template <typename... T>
void read(T &...args) {
  ((cin >> args), ...);
}
template <typename... T>
void write(T &&...args) {
  ((cout << args << endl), ...);
}
const long long int mod = 1e9 + 7, N = 1e5 + 1;
vector<pair<long long int, long long int>> moves = {
    {-1, 0}, {1, 0}, {0, -1}, {0, 1}};
void solve() {
  long long int n, k;
  cin >> n >> k;
  k++;
  long long int a[n];
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long int ans = 0;
  for (long long int i = 0; i < n - 1; i++) {
    long long int noe = pow(10, a[i + 1] - a[i]) - 1;
    if (k > noe) {
      ans += noe * pow(10, a[i]);
      k -= noe;
    } else {
      ans += k * pow(10, a[i]);
      k = 0;
      break;
    }
  }
  ans += k * pow(10, a[n - 1]);
  cout << ans << endl;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```