#include <bits/stdc++.h>
using namespace std;
long long T, n, k;
long long a[100], s[100], pre[18];
signed main() {
  scanf("%lld", &T);
  pre[0] = 1;
  for (long long i = 1; i <= 18; i++) pre[i] = pre[i - 1] * 10;
  while (T--) {
    scanf("%lld%lld", &n, &k);
    k++;
    for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]), s[i] = 0;
    for (long long i = 1; i <= n - 1; i++) {
      long long t = pre[a[i + 1] - a[i]] - 1;
      if (t < k) {
        s[i] = t;
        k -= t;
      } else {
        s[i] = k;
        k = 0;
        break;
      }
    }
    if (k) s[n] = k;
    long long ans = 0;
    for (long long i = 1; i <= n; i++) ans += s[i] * pre[a[i]];
    printf("%lld\n", ans);
  }
}

### 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;
long long T, n, k;
long long a[100], s[100], pre[18];
signed main() {
  scanf("%lld", &T);
  pre[0] = 1;
  for (long long i = 1; i <= 18; i++) pre[i] = pre[i - 1] * 10;
  while (T--) {
    scanf("%lld%lld", &n, &k);
    k++;
    for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]), s[i] = 0;
    for (long long i = 1; i <= n - 1; i++) {
      long long t = pre[a[i + 1] - a[i]] - 1;
      if (t < k) {
        s[i] = t;
        k -= t;
      } else {
        s[i] = k;
        k = 0;
        break;
      }
    }
    if (k) s[n] = k;
    long long ans = 0;
    for (long long i = 1; i <= n; i++) ans += s[i] * pre[a[i]];
    printf("%lld\n", ans);
  }
}

```