#include <bits/stdc++.h>
using namespace std;
using Vi = vector<int>;
using i64 = long long;
using ll = long long;
using Pi = pair<int, int>;
mt19937 mrand(time(0));
int rnd(int x) { return mrand() % x; }
const int N = 500005;
const int M = 40005;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n;
    ll k;
    cin >> n >> k;
    ++k;
    Vi a(n);
    vector<ll> ten(15);
    ten[0] = 1;
    for (int i = 1; i < 15; ++i) ten[i] = ten[i - 1] * 10;
    for (int i = 0; i < n; ++i) cin >> a[i];
    ll ans = 0;
    for (int i = 0; i < n; ++i) {
      ll t = (i == n - 1 ? 1000000001 : ten[a[i + 1] - a[i]] - 1);
      ll tmd = min(t, k);
      ans += tmd * ten[a[i]];
      k -= tmd;
    }
    cout << ans << endl;
  }
  return 0;
}

### 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;
using Vi = vector<int>;
using i64 = long long;
using ll = long long;
using Pi = pair<int, int>;
mt19937 mrand(time(0));
int rnd(int x) { return mrand() % x; }
const int N = 500005;
const int M = 40005;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n;
    ll k;
    cin >> n >> k;
    ++k;
    Vi a(n);
    vector<ll> ten(15);
    ten[0] = 1;
    for (int i = 1; i < 15; ++i) ten[i] = ten[i - 1] * 10;
    for (int i = 0; i < n; ++i) cin >> a[i];
    ll ans = 0;
    for (int i = 0; i < n; ++i) {
      ll t = (i == n - 1 ? 1000000001 : ten[a[i + 1] - a[i]] - 1);
      ll tmd = min(t, k);
      ans += tmd * ten[a[i]];
      k -= tmd;
    }
    cout << ans << endl;
  }
  return 0;
}

```