#include <bits/stdc++.h>
using namespace std;
long long n, k;
long long a[15];
long long dp[15];
long long ans[15];
string an;
signed main() {
  ios::sync_with_stdio(0);
  long long T;
  cin >> T;
  while (T--) {
    memset(dp, 0, sizeof dp);
    memset(ans, 0, sizeof ans);
    memset(a, 0, sizeof a);
    cin >> n >> k;
    for (long long i = 1; i <= n; i++) {
      cin >> a[i];
    }
    for (long long i = 1; i <= n - 1; i++) {
      dp[i] = pow(10ll, a[i + 1] - a[i]) - 1ll;
    }
    long long so = k + 1;
    for (long long i = 1; i <= n - 1; i++) {
      if (so >= dp[i]) {
        so -= dp[i];
        ans[i] = dp[i];
      } else {
        ans[i] = so;
        so = 0;
        break;
      }
    }
    an = "";
    for (long long i = 1; i < n; i++) {
      if (ans[i] == 0) break;
      string t = to_string(ans[i]);
      reverse(t.begin(), t.end());
      an += t;
    }
    if (so > 0) {
      string t = to_string(so);
      reverse(t.begin(), t.end());
      an += t;
    }
    reverse(an.begin(), an.end());
    cout << an << endl;
  }
}

### 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;
long long n, k;
long long a[15];
long long dp[15];
long long ans[15];
string an;
signed main() {
  ios::sync_with_stdio(0);
  long long T;
  cin >> T;
  while (T--) {
    memset(dp, 0, sizeof dp);
    memset(ans, 0, sizeof ans);
    memset(a, 0, sizeof a);
    cin >> n >> k;
    for (long long i = 1; i <= n; i++) {
      cin >> a[i];
    }
    for (long long i = 1; i <= n - 1; i++) {
      dp[i] = pow(10ll, a[i + 1] - a[i]) - 1ll;
    }
    long long so = k + 1;
    for (long long i = 1; i <= n - 1; i++) {
      if (so >= dp[i]) {
        so -= dp[i];
        ans[i] = dp[i];
      } else {
        ans[i] = so;
        so = 0;
        break;
      }
    }
    an = "";
    for (long long i = 1; i < n; i++) {
      if (ans[i] == 0) break;
      string t = to_string(ans[i]);
      reverse(t.begin(), t.end());
      an += t;
    }
    if (so > 0) {
      string t = to_string(so);
      reverse(t.begin(), t.end());
      an += t;
    }
    reverse(an.begin(), an.end());
    cout << an << endl;
  }
}

```