#include <bits/stdc++.h>
using namespace std;
long long maxn = 2000001;
long long mod = 1000000007;
vector<long long> sieveVector(maxn + 1, 1);
vector<long long> factVector(maxn + 1, 1);
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b % 2)
      res = res * a, b--;
    else
      a = a * a, b /= 2;
  }
  return res;
}
long long powerMod(long long a, long long b, long long m) {
  long long res = 1;
  while (b) {
    if (b % 2)
      res = (res % m * a % m) % m, b--;
    else
      a = (a % m * a % m) % m, b /= 2;
  }
  return res;
}
long long modIn(long long n, long long m) { return powerMod(n, m - 2, m); }
void factorial() {
  for (long long i = 1; i <= maxn; i++)
    factVector[i] = (i % mod * factVector[i - 1] % mod);
}
void sieve() {
  for (long long i = 2; i * i <= maxn; i++) {
    if (sieveVector[i]) {
      for (long long j = i * i; j <= maxn; j += i) {
        sieveVector[j] = 0;
      }
    }
  }
}
long long ncr(long long n, long long r) {
  if (r > n) return 0;
  return (factVector[n] % mod *
          modIn((factVector[n - r] % mod * factVector[r] % mod) % mod, mod)) %
         mod;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> v(n);
  for (long long I = 0; I < n; I++) cin >> v[I];
  ;
  long long ans = 0;
  k++;
  for (long long i = 0; i < n - 1; i++) {
    if (!k) break;
    long long d = power(10, v[i + 1]) - 1;
    long long maxe = d / power(10, v[i]);
    ans += min(k, maxe) * power(10, v[i]);
    k -= min(k, maxe);
  }
  ans += (k * power(10, v.back()));
  cout << ans << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long tc;
  cin >> tc;
  while (tc--) solve();
}

### 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;
long long maxn = 2000001;
long long mod = 1000000007;
vector<long long> sieveVector(maxn + 1, 1);
vector<long long> factVector(maxn + 1, 1);
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b % 2)
      res = res * a, b--;
    else
      a = a * a, b /= 2;
  }
  return res;
}
long long powerMod(long long a, long long b, long long m) {
  long long res = 1;
  while (b) {
    if (b % 2)
      res = (res % m * a % m) % m, b--;
    else
      a = (a % m * a % m) % m, b /= 2;
  }
  return res;
}
long long modIn(long long n, long long m) { return powerMod(n, m - 2, m); }
void factorial() {
  for (long long i = 1; i <= maxn; i++)
    factVector[i] = (i % mod * factVector[i - 1] % mod);
}
void sieve() {
  for (long long i = 2; i * i <= maxn; i++) {
    if (sieveVector[i]) {
      for (long long j = i * i; j <= maxn; j += i) {
        sieveVector[j] = 0;
      }
    }
  }
}
long long ncr(long long n, long long r) {
  if (r > n) return 0;
  return (factVector[n] % mod *
          modIn((factVector[n - r] % mod * factVector[r] % mod) % mod, mod)) %
         mod;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> v(n);
  for (long long I = 0; I < n; I++) cin >> v[I];
  ;
  long long ans = 0;
  k++;
  for (long long i = 0; i < n - 1; i++) {
    if (!k) break;
    long long d = power(10, v[i + 1]) - 1;
    long long maxe = d / power(10, v[i]);
    ans += min(k, maxe) * power(10, v[i]);
    k -= min(k, maxe);
  }
  ans += (k * power(10, v.back()));
  cout << ans << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long tc;
  cin >> tc;
  while (tc--) solve();
}

```