#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
int T;
long long n, k;
long long cost[20], b[20], a[20];
int main() {
  scanf("%d", &T);
  b[0] = 1;
  for (int i = 1; i <= 10; ++i) {
    b[i] = b[i - 1] * 10;
  }
  while (T--) {
    scanf("%lld%lld", &n, &k);
    k++;
    memset(cost, 0, sizeof(cost));
    memset(a, 0, sizeof(a));
    cost[0] = 1;
    for (int i = 1; i <= n; ++i) {
      int bits;
      scanf("%d", &bits);
      cost[bits] = 1;
    }
    for (int i = 1; i <= 10; ++i) {
      if (!cost[i]) {
        cost[i] = cost[i - 1] * 10;
      }
    }
    long long ans = 0, tot = 0;
    for (int i = 0; i <= 9; ++i) {
      k -= 9 * cost[i];
      a[i] = 9;
      ans += 9 * b[i];
      if (k <= 0) break;
    }
    if (k > 0) {
      ans += (k / cost[10]) * b[10];
      a[10] += (k / cost[10]);
      if (k % cost[10]) {
        ans += b[10];
        a[10]++;
      }
      k -= cost[10] * a[10];
    }
    for (int i = 10; i >= 0; --i) {
      while (a[i] != 0 && (-k) >= cost[i]) {
        ans -= b[i];
        a[i]--;
        k += cost[i];
      }
    }
    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 mod = 1e9 + 7;
int T;
long long n, k;
long long cost[20], b[20], a[20];
int main() {
  scanf("%d", &T);
  b[0] = 1;
  for (int i = 1; i <= 10; ++i) {
    b[i] = b[i - 1] * 10;
  }
  while (T--) {
    scanf("%lld%lld", &n, &k);
    k++;
    memset(cost, 0, sizeof(cost));
    memset(a, 0, sizeof(a));
    cost[0] = 1;
    for (int i = 1; i <= n; ++i) {
      int bits;
      scanf("%d", &bits);
      cost[bits] = 1;
    }
    for (int i = 1; i <= 10; ++i) {
      if (!cost[i]) {
        cost[i] = cost[i - 1] * 10;
      }
    }
    long long ans = 0, tot = 0;
    for (int i = 0; i <= 9; ++i) {
      k -= 9 * cost[i];
      a[i] = 9;
      ans += 9 * b[i];
      if (k <= 0) break;
    }
    if (k > 0) {
      ans += (k / cost[10]) * b[10];
      a[10] += (k / cost[10]);
      if (k % cost[10]) {
        ans += b[10];
        a[10]++;
      }
      k -= cost[10] * a[10];
    }
    for (int i = 10; i >= 0; --i) {
      while (a[i] != 0 && (-k) >= cost[i]) {
        ans -= b[i];
        a[i]--;
        k += cost[i];
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```