#include <bits/stdc++.h>
using namespace std;
void in(int &n) {
  n = 0;
  int f = 1;
  char c = getchar();
  while (c < '0' || '9' < c) {
    if (c == '-') f *= -1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') {
    n = n * 10 + (c - '0');
    c = getchar();
  }
  n *= f;
}
void llin(long long &n) {
  n = 0;
  long long f = 1;
  char c = getchar();
  while (c < '0' || '9' < c) {
    if (c == '-') f *= -1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') {
    n = n * 10 + (c - '0');
    c = getchar();
  }
  n *= f;
}
long long pw10[25], a[25], b[25], sum[25];
int main() {
  pw10[0] = 1;
  for (int i = 1; i <= 18; i++) pw10[i] = pw10[i - 1] * 10ll;
  int t;
  in(t);
  while (t--) {
    memset(a, 0, sizeof(a));
    memset(b, 0, sizeof(b));
    memset(sum, 0, sizeof(sum));
    int n;
    long long m;
    in(n);
    llin(m);
    m++;
    for (int i = 1; i <= n; i++) {
      int x;
      in(x);
      a[x] = 1;
      for (int j = x + 1; j <= 18; j++) a[j] = a[j - 1] * 10ll;
    }
    sum[0] = a[0] * 9ll;
    for (int i = 1; i <= 18; i++) sum[i] = sum[i - 1] + 9ll * a[i];
    while (m) {
      if (m <= sum[0]) {
        long long tmp = m;
        tmp = tmp / a[0] + ((tmp % a[0]) != 0);
        b[0] = tmp;
        m -= tmp * a[0];
        continue;
      }
      for (int i = 0; i < 18; i++) {
        if (sum[i] <= m && m <= sum[i + 1]) {
          long long tmp = m - sum[i];
          tmp = tmp / a[i + 1] + ((tmp % a[i + 1]) != 0);
          b[i + 1] = tmp;
          m -= tmp * a[i + 1];
          break;
        }
      }
    }
    long long ans = 0;
    for (int i = 18; i >= 0; i--) ans = ans * 10ll + b[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;
void in(int &n) {
  n = 0;
  int f = 1;
  char c = getchar();
  while (c < '0' || '9' < c) {
    if (c == '-') f *= -1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') {
    n = n * 10 + (c - '0');
    c = getchar();
  }
  n *= f;
}
void llin(long long &n) {
  n = 0;
  long long f = 1;
  char c = getchar();
  while (c < '0' || '9' < c) {
    if (c == '-') f *= -1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') {
    n = n * 10 + (c - '0');
    c = getchar();
  }
  n *= f;
}
long long pw10[25], a[25], b[25], sum[25];
int main() {
  pw10[0] = 1;
  for (int i = 1; i <= 18; i++) pw10[i] = pw10[i - 1] * 10ll;
  int t;
  in(t);
  while (t--) {
    memset(a, 0, sizeof(a));
    memset(b, 0, sizeof(b));
    memset(sum, 0, sizeof(sum));
    int n;
    long long m;
    in(n);
    llin(m);
    m++;
    for (int i = 1; i <= n; i++) {
      int x;
      in(x);
      a[x] = 1;
      for (int j = x + 1; j <= 18; j++) a[j] = a[j - 1] * 10ll;
    }
    sum[0] = a[0] * 9ll;
    for (int i = 1; i <= 18; i++) sum[i] = sum[i - 1] + 9ll * a[i];
    while (m) {
      if (m <= sum[0]) {
        long long tmp = m;
        tmp = tmp / a[0] + ((tmp % a[0]) != 0);
        b[0] = tmp;
        m -= tmp * a[0];
        continue;
      }
      for (int i = 0; i < 18; i++) {
        if (sum[i] <= m && m <= sum[i + 1]) {
          long long tmp = m - sum[i];
          tmp = tmp / a[i + 1] + ((tmp % a[i + 1]) != 0);
          b[i + 1] = tmp;
          m -= tmp * a[i + 1];
          break;
        }
      }
    }
    long long ans = 0;
    for (int i = 18; i >= 0; i--) ans = ans * 10ll + b[i];
    printf("%lld\n", ans);
  }
  return 0;
}

```