#include <bits/stdc++.h>
using namespace std;
const int mo = 1e9 + 7;
long long int a[100];
long long fastPower(long long base, long long power) {
  long long result = 1;
  while (power > 0) {
    if (power & 1) {
      result = result * base;
    }
    power >>= 1;
    base = (base * base);
  }
  return result;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    unsigned long long int n, k, ans = 0;
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    bool flag = false;
    for (int i = 1; i <= n; i++) {
      if (i == 1) {
        if (a[i] != 0) {
          ans = 1;
          flag = true;
          break;
        }
      } else {
        if (k < fastPower(10ll, a[i] - a[i - 1]) - 1) {
          ans += (k + 1) * fastPower(10ll, a[i - 1]);
          flag = true;
          break;
        } else {
          for (int j = a[i - 1]; j < a[i]; j++) {
            ans += fastPower(10ll, j) * 9;
          }
          k = k - fastPower(10ll, a[i] - a[i - 1]) + 1;
        }
      }
    }
    if (flag) {
      cout << ans << "\n";
      ;
    } else {
      ans += (k + 1) * fastPower(10ll, a[n]);
      cout << ans << "\n";
      ;
    }
  }
  return 0;
}

### Prompt

Please formulate 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 mo = 1e9 + 7;
long long int a[100];
long long fastPower(long long base, long long power) {
  long long result = 1;
  while (power > 0) {
    if (power & 1) {
      result = result * base;
    }
    power >>= 1;
    base = (base * base);
  }
  return result;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    unsigned long long int n, k, ans = 0;
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
    }
    bool flag = false;
    for (int i = 1; i <= n; i++) {
      if (i == 1) {
        if (a[i] != 0) {
          ans = 1;
          flag = true;
          break;
        }
      } else {
        if (k < fastPower(10ll, a[i] - a[i - 1]) - 1) {
          ans += (k + 1) * fastPower(10ll, a[i - 1]);
          flag = true;
          break;
        } else {
          for (int j = a[i - 1]; j < a[i]; j++) {
            ans += fastPower(10ll, j) * 9;
          }
          k = k - fastPower(10ll, a[i] - a[i - 1]) + 1;
        }
      }
    }
    if (flag) {
      cout << ans << "\n";
      ;
    } else {
      ans += (k + 1) * fastPower(10ll, a[n]);
      cout << ans << "\n";
      ;
    }
  }
  return 0;
}

```