#include <bits/stdc++.h>
using namespace std;
using ll = int64_t;
using ull = uint64_t;
using ld = long double;
using vi = vector<int64_t>;
using pii = pair<int64_t, int64_t>;
ll powr(ll a, ll b) {
  ll res = 1;
  while (b) {
    if (b & 1) res = (res * a);
    a = (a * a);
    b >>= 1;
  }
  return res;
}
void solve() {
  int64_t n, k;
  cin >> n >> k;
  k++;
  int64_t a[n];
  for (int64_t i = 0; i < n; i++) {
    cin >> a[i];
    a[i] = powr(10, a[i]);
  }
  sort(a, a + n);
  int64_t ans = 0;
  for (int64_t i = 0; i < n - 1 && k > 0; i++) {
    int64_t den = min(k, (a[i + 1] / a[i]) - 1);
    ans += a[i] * den;
    k -= den;
  }
  if (k > 0) {
    ans += a[n - 1] * k;
  }
  cout << ans << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int64_t t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Your task is to create 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;
using ll = int64_t;
using ull = uint64_t;
using ld = long double;
using vi = vector<int64_t>;
using pii = pair<int64_t, int64_t>;
ll powr(ll a, ll b) {
  ll res = 1;
  while (b) {
    if (b & 1) res = (res * a);
    a = (a * a);
    b >>= 1;
  }
  return res;
}
void solve() {
  int64_t n, k;
  cin >> n >> k;
  k++;
  int64_t a[n];
  for (int64_t i = 0; i < n; i++) {
    cin >> a[i];
    a[i] = powr(10, a[i]);
  }
  sort(a, a + n);
  int64_t ans = 0;
  for (int64_t i = 0; i < n - 1 && k > 0; i++) {
    int64_t den = min(k, (a[i + 1] / a[i]) - 1);
    ans += a[i] * den;
    k -= den;
  }
  if (k > 0) {
    ans += a[n - 1] * k;
  }
  cout << ans << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int64_t t;
  cin >> t;
  while (t--) solve();
}

```