#include <bits/stdc++.h>
using namespace std;
long long a[5007], b[5007];
long long pre[5007], back[5007];
long long ksm(long long base, long long p) {
  long long r = 1;
  while (p > 0) {
    if (p & 1) {
      r = r * base;
    }
    p >>= 1;
    base = (base * base);
  }
  return r;
}
int main() {
  ios::sync_with_stdio(false);
  long long T;
  cin >> T;
  while (T--) {
    long long n, m;
    cin >> n >> m;
    for (long long i = 1; i <= n; i++) {
      long long x;
      cin >> x;
      a[i] = ksm(10, x);
      b[i] = x;
    }
    long long ans = 0;
    long long t = m + 1;
    for (long long i = 1; i < n; i++) {
      long long z = b[i + 1] - b[i];
      long long q = ksm(10, z) - 1;
      if (t - q >= 0) {
        t = t - q;
        ans += q * a[i];
      } else {
        ans += t * a[i];
        t = 0;
        break;
      }
    }
    if (t != 0) ans += t * a[n];
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 a[5007], b[5007];
long long pre[5007], back[5007];
long long ksm(long long base, long long p) {
  long long r = 1;
  while (p > 0) {
    if (p & 1) {
      r = r * base;
    }
    p >>= 1;
    base = (base * base);
  }
  return r;
}
int main() {
  ios::sync_with_stdio(false);
  long long T;
  cin >> T;
  while (T--) {
    long long n, m;
    cin >> n >> m;
    for (long long i = 1; i <= n; i++) {
      long long x;
      cin >> x;
      a[i] = ksm(10, x);
      b[i] = x;
    }
    long long ans = 0;
    long long t = m + 1;
    for (long long i = 1; i < n; i++) {
      long long z = b[i + 1] - b[i];
      long long q = ksm(10, z) - 1;
      if (t - q >= 0) {
        t = t - q;
        ans += q * a[i];
      } else {
        ans += t * a[i];
        t = 0;
        break;
      }
    }
    if (t != 0) ans += t * a[n];
    cout << ans << endl;
  }
  return 0;
}

```