#include <bits/stdc++.h>
using namespace std;
const long long maxn = 11;
long long a[maxn];
long long b[maxn];
inline long long qpow(long long x, long long y) {
  long long res(1);
  while (y) {
    if (y & 1) res = 1ll * res * x;
    x = 1ll * x * x;
    y >>= 1;
  }
  return res;
}
signed main() {
  ios::sync_with_stdio(false);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    k++;
    for (long long i = 1; i <= n; i++) cin >> a[i];
    for (long long i = 2; i <= n; i++) b[i] = a[i] - a[i - 1];
    long long now = 0;
    for (long long i = 2; i <= n; i++) {
      long long tmp = qpow(10, b[i]) - 1;
      if (k > tmp) {
        k -= tmp;
        now += tmp * qpow(10, a[i - 1]);
      } else {
        now += k * qpow(10, a[i - 1]);
        k = 0;
        break;
      }
    }
    if (k > 0) {
      now += k * qpow(10, a[n]);
    }
    cout << now << "\n";
  }
}

### Prompt

Generate a CPP solution 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;
const long long maxn = 11;
long long a[maxn];
long long b[maxn];
inline long long qpow(long long x, long long y) {
  long long res(1);
  while (y) {
    if (y & 1) res = 1ll * res * x;
    x = 1ll * x * x;
    y >>= 1;
  }
  return res;
}
signed main() {
  ios::sync_with_stdio(false);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    k++;
    for (long long i = 1; i <= n; i++) cin >> a[i];
    for (long long i = 2; i <= n; i++) b[i] = a[i] - a[i - 1];
    long long now = 0;
    for (long long i = 2; i <= n; i++) {
      long long tmp = qpow(10, b[i]) - 1;
      if (k > tmp) {
        k -= tmp;
        now += tmp * qpow(10, a[i - 1]);
      } else {
        now += k * qpow(10, a[i - 1]);
        k = 0;
        break;
      }
    }
    if (k > 0) {
      now += k * qpow(10, a[n]);
    }
    cout << now << "\n";
  }
}

```