#include <bits/stdc++.h>
using namespace std;
const double Pi = acos(-1.0);
const long long dx[] = {0, 0, 1, -1};
const long long dy[] = {1, -1, 0, 0};
const long long N = (long long)1e6 + 17;
const long long M = (long long)2e3 + 69;
const long long inf = (long long)1e14 + 3;
const long long mod = (long long)1e9 + 7;
long long sq(long long x) { return x * x; }
long long t = 1, a[N];
void solve() {
  long long n, k, cnt = 0;
  cin >> n >> k;
  k++;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (long long i = 2; i <= n; i++) {
    long long x = pow(10, a[i] - a[i - 1]) - 1;
    if (k >= x)
      k -= x, cnt += a[i] - a[i - 1];
    else
      break;
  }
  if (k) cout << k;
  while (cnt--) cout << 9;
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Generate 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 double Pi = acos(-1.0);
const long long dx[] = {0, 0, 1, -1};
const long long dy[] = {1, -1, 0, 0};
const long long N = (long long)1e6 + 17;
const long long M = (long long)2e3 + 69;
const long long inf = (long long)1e14 + 3;
const long long mod = (long long)1e9 + 7;
long long sq(long long x) { return x * x; }
long long t = 1, a[N];
void solve() {
  long long n, k, cnt = 0;
  cin >> n >> k;
  k++;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (long long i = 2; i <= n; i++) {
    long long x = pow(10, a[i] - a[i - 1]) - 1;
    if (k >= x)
      k -= x, cnt += a[i] - a[i - 1];
    else
      break;
  }
  if (k) cout << k;
  while (cnt--) cout << 9;
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```