#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1);
long long q_pow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
const int N = 2e5 + 456;
long long a[100], dp[12], b[100];
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
      int x;
      cin >> x;
      if (!x) {
        a[i] = 1;
      } else {
        long long sum = 1;
        for (int j = 1; j <= x; j++) {
          sum *= 10;
        }
        a[i] = sum;
      }
    }
    a[n + 1] = 2 * a[n];
    long long ans = 0, sum = 0;
    for (int i = 1; i <= n; i++) {
      b[i] = (a[i + 1] / a[i] - 1);
    }
    int ok = 0;
    for (int i = 1; i <= n - 1; i++) {
      if (k - ans < b[i]) {
        ok = 1;
        sum += (k - ans + 1) * a[i];
        break;
      } else {
        ans += b[i];
        sum += b[i] * a[i];
      }
    }
    if (ok)
      cout << sum << '\n';
    else {
      sum += (k - ans + 1) * a[n];
      cout << sum << '\n';
    }
  }
  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;
const double pi = acos(-1);
long long q_pow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
const int N = 2e5 + 456;
long long a[100], dp[12], b[100];
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
      int x;
      cin >> x;
      if (!x) {
        a[i] = 1;
      } else {
        long long sum = 1;
        for (int j = 1; j <= x; j++) {
          sum *= 10;
        }
        a[i] = sum;
      }
    }
    a[n + 1] = 2 * a[n];
    long long ans = 0, sum = 0;
    for (int i = 1; i <= n; i++) {
      b[i] = (a[i + 1] / a[i] - 1);
    }
    int ok = 0;
    for (int i = 1; i <= n - 1; i++) {
      if (k - ans < b[i]) {
        ok = 1;
        sum += (k - ans + 1) * a[i];
        break;
      } else {
        ans += b[i];
        sum += b[i] * a[i];
      }
    }
    if (ok)
      cout << sum << '\n';
    else {
      sum += (k - ans + 1) * a[n];
      cout << sum << '\n';
    }
  }
  return 0;
}

```