#include <bits/stdc++.h>
using namespace std;
const long long inf = 0x3f3f3f3f3f3f3f3f;
template <typename T>
inline void read(T& X) {
  X = 0;
  int w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
  if (w) X = -X;
}
long long a[20], p[20];
int main() {
  int t;
  for (int i = 0; i <= 9; i++) p[i] = pow(10, i);
  scanf("%d", &t);
  while (t--) {
    long long k, n;
    long long ans = 0;
    scanf("%lld %lld", &n, &k);
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
      a[i] = p[a[i]];
    }
    for (int i = 1; i <= n; i++) {
      if (i == n) {
        ans = ans + (k + 1) * a[i];
        break;
      }
      long long now = (a[i + 1] - a[i]) / a[i];
      if (now <= k) {
        ans = ans + a[i + 1] - a[i];
        k = k - now;
      } else {
        ans = ans + a[i] * (k + 1);
        break;
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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 namespace std;
const long long inf = 0x3f3f3f3f3f3f3f3f;
template <typename T>
inline void read(T& X) {
  X = 0;
  int w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
  if (w) X = -X;
}
long long a[20], p[20];
int main() {
  int t;
  for (int i = 0; i <= 9; i++) p[i] = pow(10, i);
  scanf("%d", &t);
  while (t--) {
    long long k, n;
    long long ans = 0;
    scanf("%lld %lld", &n, &k);
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &a[i]);
      a[i] = p[a[i]];
    }
    for (int i = 1; i <= n; i++) {
      if (i == n) {
        ans = ans + (k + 1) * a[i];
        break;
      }
      long long now = (a[i + 1] - a[i]) / a[i];
      if (now <= k) {
        ans = ans + a[i + 1] - a[i];
        k = k - now;
      } else {
        ans = ans + a[i] * (k + 1);
        break;
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```