#include <bits/stdc++.h>
using namespace std;
const int maxn = 100;
long long n, k;
int a[15];
long long rc[15], p[15], ans;
void solve() {
  scanf("%d%d", &n, &k);
  k++;
  scanf("%d", &a[1]);
  rc[1] = 1;
  long long now = 1, tmp;
  int ss = 0;
  for (int i = 2; i <= n; i++) {
    scanf("%d", &a[i]);
    tmp = 1;
    while (ss < a[i]) now *= 10, tmp *= 10, ss++;
    p[i - 1] = tmp - 1;
    rc[i] = now;
  }
  ans = 0;
  for (int i = 1; i < n; i++) {
    if (k >= p[i])
      ans += p[i] * rc[i], k -= p[i];
    else {
      ans += rc[i] * k;
      k = 0;
    }
  }
  if (k) ans += rc[n] * k;
  printf("%lld\n", ans);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
  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 int maxn = 100;
long long n, k;
int a[15];
long long rc[15], p[15], ans;
void solve() {
  scanf("%d%d", &n, &k);
  k++;
  scanf("%d", &a[1]);
  rc[1] = 1;
  long long now = 1, tmp;
  int ss = 0;
  for (int i = 2; i <= n; i++) {
    scanf("%d", &a[i]);
    tmp = 1;
    while (ss < a[i]) now *= 10, tmp *= 10, ss++;
    p[i - 1] = tmp - 1;
    rc[i] = now;
  }
  ans = 0;
  for (int i = 1; i < n; i++) {
    if (k >= p[i])
      ans += p[i] * rc[i], k -= p[i];
    else {
      ans += rc[i] * k;
      k = 0;
    }
  }
  if (k) ans += rc[n] * k;
  printf("%lld\n", ans);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
  return 0;
}

```