#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e2 + 10;
long long a[maxn], bin[maxn], can[maxn];
long long n, k, ans;
signed main() {
  bin[0] = 1;
  for (register long long i = 1; i <= 9; ++i) bin[i] = bin[i - 1] * 10;
  long long T;
  cin >> T;
  while (T--) {
    cin >> n >> k;
    for (register long long i = 1; i <= n; ++i) cin >> a[i];
    for (register long long i = 1; i < n; ++i)
      can[i] = (bin[a[i + 1]] - bin[a[i]]) / bin[a[i]];
    ans = 0;
    for (register long long i = 1; i < n; ++i) {
      if (can[i] <= k) {
        k -= can[i];
        ans += can[i] * bin[a[i]];
        if (!k) ans += bin[a[i + 1]];
      } else
        ans += (k + 1) * bin[a[i]], k = 0;
      if (!k) break;
    }
    if (k) ans += bin[a[n]] * (k + 1);
    cout << ans;
    puts("");
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 long long maxn = 1e2 + 10;
long long a[maxn], bin[maxn], can[maxn];
long long n, k, ans;
signed main() {
  bin[0] = 1;
  for (register long long i = 1; i <= 9; ++i) bin[i] = bin[i - 1] * 10;
  long long T;
  cin >> T;
  while (T--) {
    cin >> n >> k;
    for (register long long i = 1; i <= n; ++i) cin >> a[i];
    for (register long long i = 1; i < n; ++i)
      can[i] = (bin[a[i + 1]] - bin[a[i]]) / bin[a[i]];
    ans = 0;
    for (register long long i = 1; i < n; ++i) {
      if (can[i] <= k) {
        k -= can[i];
        ans += can[i] * bin[a[i]];
        if (!k) ans += bin[a[i + 1]];
      } else
        ans += (k + 1) * bin[a[i]], k = 0;
      if (!k) break;
    }
    if (k) ans += bin[a[n]] * (k + 1);
    cout << ans;
    puts("");
  }
  return 0;
}

```