#include <bits/stdc++.h>
using namespace std;
const int N = 15;
int a[N];
long long ksm(long long a, long long b, long long c) {
  int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % c;
    a = (a * a) % c;
    b >>= 1;
  }
  return res % c;
}
int read() {
  int n;
  cin >> n;
  return n;
}
void solve() {
  int n;
  long long k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  int pos = 0;
  vector<long long> v;
  bool f = 0;
  for (int i = 1; i <= n; i++) {
    if (i == 1 && a[i] != 0) {
      cout << 0 << '\n';
      return;
    }
    v.push_back(min(ksm(10, a[i] - pos, 1e18) - 1, k));
    if (k + 1 == ksm(10, a[i] - pos, 1e18)) f = true;
    k = max(0ll, k - ksm(10, a[i] - pos, 1e18) + 1);
    if (k == 0) break;
    pos = a[i];
  }
  if (v.size() == 1) {
    cout << v[0] + k + 1 << '\n';
    return;
  }
  if (k != 0) {
    v.push_back(k);
    v.back()++;
    for (int i = v.size() - 1; i > 0; i--) cout << v[i];
    cout << '\n';
    return;
  }
  if (k == 0) {
    if (!f) {
      v.back()++;
    } else
      v.push_back(1);
  }
  for (int i = v.size() - 1; i > 0; i--) cout << v[i];
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int _;
  cin >> _;
  while (_--) {
    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;
const int N = 15;
int a[N];
long long ksm(long long a, long long b, long long c) {
  int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % c;
    a = (a * a) % c;
    b >>= 1;
  }
  return res % c;
}
int read() {
  int n;
  cin >> n;
  return n;
}
void solve() {
  int n;
  long long k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  int pos = 0;
  vector<long long> v;
  bool f = 0;
  for (int i = 1; i <= n; i++) {
    if (i == 1 && a[i] != 0) {
      cout << 0 << '\n';
      return;
    }
    v.push_back(min(ksm(10, a[i] - pos, 1e18) - 1, k));
    if (k + 1 == ksm(10, a[i] - pos, 1e18)) f = true;
    k = max(0ll, k - ksm(10, a[i] - pos, 1e18) + 1);
    if (k == 0) break;
    pos = a[i];
  }
  if (v.size() == 1) {
    cout << v[0] + k + 1 << '\n';
    return;
  }
  if (k != 0) {
    v.push_back(k);
    v.back()++;
    for (int i = v.size() - 1; i > 0; i--) cout << v[i];
    cout << '\n';
    return;
  }
  if (k == 0) {
    if (!f) {
      v.back()++;
    } else
      v.push_back(1);
  }
  for (int i = v.size() - 1; i > 0; i--) cout << v[i];
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int _;
  cin >> _;
  while (_--) {
    solve();
  }
  return 0;
}

```