#include <bits/stdc++.h>
using std::min;
int a[11], p10[11];
int main() {
  int T, n, k;
  long long res;
  scanf("%d", &T);
  for (int i = 0, j = 1; i <= 9; ++i, j *= 10) p10[i] = j;
  while (T--) {
    res = 0;
    scanf("%d%d", &n, &k);
    ++k;
    for (int i = 1; i <= n; ++i) scanf("%d", a + i);
    for (int i = 1; i < n; ++i) {
      int max_use = p10[a[i + 1] - a[i]] - 1;
      int use = min(k, max_use);
      res += 1ll * use * p10[a[i]];
      if ((k -= use) == 0) break;
    }
    if (k) res += 1ll * p10[a[n]] * k;
    printf("%lld\n", res);
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 std::min;
int a[11], p10[11];
int main() {
  int T, n, k;
  long long res;
  scanf("%d", &T);
  for (int i = 0, j = 1; i <= 9; ++i, j *= 10) p10[i] = j;
  while (T--) {
    res = 0;
    scanf("%d%d", &n, &k);
    ++k;
    for (int i = 1; i <= n; ++i) scanf("%d", a + i);
    for (int i = 1; i < n; ++i) {
      int max_use = p10[a[i + 1] - a[i]] - 1;
      int use = min(k, max_use);
      res += 1ll * use * p10[a[i]];
      if ((k -= use) == 0) break;
    }
    if (k) res += 1ll * p10[a[n]] * k;
    printf("%lld\n", res);
  }
  return 0;
}

```