#include <bits/stdc++.h>
using ll = long long;
using namespace std;
const int maxn = 1e3 + 5;
ll ksm(ll a, ll n) {
  ll ans = 1;
  while (n) {
    if (n % 2) ans = (a * ans);
    a = (a * a);
    n /= 2;
  }
  return ans;
}
void solve() {
  ll a[maxn], dp[maxn], ten[maxn], n, k, now;
  for (int i = 0; i <= 10; i++) {
    ten[i] = ksm(10, i) - 1;
  }
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  now = n + 1;
  for (int i = 2; i <= n; i++) {
    dp[i] = ten[a[i] - a[i - 1]] + dp[i - 1];
    if (k < dp[i]) {
      now = i;
      break;
    }
  }
  cout << (ten[a[now - 1]] + 1) * (k - dp[now - 1] + 2) - 1 << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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 ll = long long;
using namespace std;
const int maxn = 1e3 + 5;
ll ksm(ll a, ll n) {
  ll ans = 1;
  while (n) {
    if (n % 2) ans = (a * ans);
    a = (a * a);
    n /= 2;
  }
  return ans;
}
void solve() {
  ll a[maxn], dp[maxn], ten[maxn], n, k, now;
  for (int i = 0; i <= 10; i++) {
    ten[i] = ksm(10, i) - 1;
  }
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  now = n + 1;
  for (int i = 2; i <= n; i++) {
    dp[i] = ten[a[i] - a[i - 1]] + dp[i - 1];
    if (k < dp[i]) {
      now = i;
      break;
    }
  }
  cout << (ten[a[now - 1]] + 1) * (k - dp[now - 1] + 2) - 1 << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```