#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int N = 1e5 + 6;
long long int binpower(long long int a, long long int b, long long int m) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b = b >> 1;
  }
  return res;
}
void solve() {
  int t;
  cin >> t;
  while (t--) {
    long long int n, k;
    cin >> n >> k;
    vector<int> arr(n);
    for (auto& i : arr) cin >> i;
    long long int sum = 0;
    sort(arr.begin(), arr.end());
    k++;
    for (int i = 0; i < n - 1; i++) {
      if (k == 0) break;
      int nextval = arr[i + 1];
      long long int poss = (binpower(10LL, nextval, 1e18) - sum - 1) /
                           binpower(10LL, arr[i], 1e18);
      poss = min(k, poss);
      k -= poss;
      sum += poss * (binpower(10, arr[i], 1e18));
    }
    if (k > 0) {
      sum += k * (binpower(10, arr[n - 1], 1e18));
    }
    cout << sum << endl;
  }
}
int32_t main(int argc, char* argv[]) { solve(); }

### 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 int mod = 1e9 + 7;
const int N = 1e5 + 6;
long long int binpower(long long int a, long long int b, long long int m) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b = b >> 1;
  }
  return res;
}
void solve() {
  int t;
  cin >> t;
  while (t--) {
    long long int n, k;
    cin >> n >> k;
    vector<int> arr(n);
    for (auto& i : arr) cin >> i;
    long long int sum = 0;
    sort(arr.begin(), arr.end());
    k++;
    for (int i = 0; i < n - 1; i++) {
      if (k == 0) break;
      int nextval = arr[i + 1];
      long long int poss = (binpower(10LL, nextval, 1e18) - sum - 1) /
                           binpower(10LL, arr[i], 1e18);
      poss = min(k, poss);
      k -= poss;
      sum += poss * (binpower(10, arr[i], 1e18));
    }
    if (k > 0) {
      sum += k * (binpower(10, arr[n - 1], 1e18));
    }
    cout << sum << endl;
  }
}
int32_t main(int argc, char* argv[]) { solve(); }

```