#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T MAX(T a, T b) {
  if (a > b) return a;
  return b;
}
template <class T>
inline T MIN(T a, T b) {
  if (a < b) return a;
  return b;
}
template <class T>
inline T ABS(T x) {
  if (x < 0) return -x;
  return x;
}
inline void OPEN(const string &s) {
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
const static int INF = 2123123123;
int tc;
long long n, k;
long long a[12];
long long powI(long long a, long long b) {
  long long result = 1;
  for (int(i) = (0); (i) < (b); ++(i)) {
    result *= a;
  }
  return result;
}
int main() {
  scanf("%d", &tc);
  while (tc--) {
    scanf("%lld%lld", &n, &k);
    for (int(i) = (0); (i) < (n); ++(i)) {
      scanf("%lld", &a[i]);
    }
    long long result = 0;
    long long sisaK = k;
    for (int i = 0; i < n; i++) {
      if (i + 1 < n) {
        long long maxTime = powI(10, a[i + 1] - a[i]) - 1;
        long long times = min(maxTime, sisaK);
        sisaK -= times;
        result += powI(10, a[i]) * times;
        if (sisaK == 0 && powI(10, a[i]) * (times + 1) != powI(10, a[i + 1])) {
          result += powI(10, a[i]);
          break;
        }
      } else {
        result += powI(10, a[i]) * (sisaK + 1);
      }
    }
    printf("%lld\n", result);
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
template <class T>
inline T MAX(T a, T b) {
  if (a > b) return a;
  return b;
}
template <class T>
inline T MIN(T a, T b) {
  if (a < b) return a;
  return b;
}
template <class T>
inline T ABS(T x) {
  if (x < 0) return -x;
  return x;
}
inline void OPEN(const string &s) {
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
const static int INF = 2123123123;
int tc;
long long n, k;
long long a[12];
long long powI(long long a, long long b) {
  long long result = 1;
  for (int(i) = (0); (i) < (b); ++(i)) {
    result *= a;
  }
  return result;
}
int main() {
  scanf("%d", &tc);
  while (tc--) {
    scanf("%lld%lld", &n, &k);
    for (int(i) = (0); (i) < (n); ++(i)) {
      scanf("%lld", &a[i]);
    }
    long long result = 0;
    long long sisaK = k;
    for (int i = 0; i < n; i++) {
      if (i + 1 < n) {
        long long maxTime = powI(10, a[i + 1] - a[i]) - 1;
        long long times = min(maxTime, sisaK);
        sisaK -= times;
        result += powI(10, a[i]) * times;
        if (sisaK == 0 && powI(10, a[i]) * (times + 1) != powI(10, a[i + 1])) {
          result += powI(10, a[i]);
          break;
        }
      } else {
        result += powI(10, a[i]) * (sisaK + 1);
      }
    }
    printf("%lld\n", result);
  }
  return 0;
}

```