#include <bits/stdc++.h>
using namespace std;
const int N = 20;
int T, n, k;
long long inf = 1e18;
vector<int> v;
void _clear() { v.clear(); }
long long po(long long x, long long os) {
  if (os == 0) return 1;
  long long z = po(x, os / 2);
  if (os & 1) return z * z * x;
  return z * z;
}
long long calc(int k) {
  long long ret = 0;
  long long num = 0;
  for (int i = 0; i <= n - 2; i++) {
    long long moves = min(1LL * v[i + 1] / v[i] - 1, 1LL * k - num);
    ret += moves * v[i];
    num += moves;
  }
  ret += (k - num) * v.back();
  return ret;
}
int main() {
  scanf("%d", &T);
  while (T--) {
    _clear();
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++) {
      int x;
      scanf("%d", &x);
      v.push_back(po(10, x));
    }
    sort(v.begin(), v.end());
    printf("%lld\n", calc(k + 1));
  }
}

### Prompt

Please create a solution in Cpp 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 = 20;
int T, n, k;
long long inf = 1e18;
vector<int> v;
void _clear() { v.clear(); }
long long po(long long x, long long os) {
  if (os == 0) return 1;
  long long z = po(x, os / 2);
  if (os & 1) return z * z * x;
  return z * z;
}
long long calc(int k) {
  long long ret = 0;
  long long num = 0;
  for (int i = 0; i <= n - 2; i++) {
    long long moves = min(1LL * v[i + 1] / v[i] - 1, 1LL * k - num);
    ret += moves * v[i];
    num += moves;
  }
  ret += (k - num) * v.back();
  return ret;
}
int main() {
  scanf("%d", &T);
  while (T--) {
    _clear();
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++) {
      int x;
      scanf("%d", &x);
      v.push_back(po(10, x));
    }
    sort(v.begin(), v.end());
    printf("%lld\n", calc(k + 1));
  }
}

```