#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
inline bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.second == b.second) return a.first < b.first;
  return a.second > b.second;
}
int drx[8] = {-2, -2, -1, -1, 1, 1, 2, 2};
int dcy[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
int dirx[4] = {-1, 0, 1, 0};
int diry[4] = {0, -1, 0, 1};
long long binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long InverseMod(long long base, long long pow) {
  if (pow == 0) return 1;
  long long ans = InverseMod(base, pow / 2);
  ans = (ans * ans) % 1000000007;
  if (pow & 1) {
    return (ans * base) % 1000000007;
  } else {
    return ans;
  }
}
bool isprime(long long num) {
  if (num < 2) return false;
  for (long long i = 2; i * i <= num; i++) {
    if (num % i == 0) return false;
  }
  return true;
}
long long EularPHI(long long num) {
  double ans = num;
  for (long long i = 2; i * i <= num; i++) {
    if (num % i == 0) {
      while (num % i == 0) {
        num /= i;
      }
      ans *= (1.0 - (1.0 / (double)i));
    }
  }
  if (num > 1) ans *= (1.0 - (1.0 / (double)num));
  return (long long)ans;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
  return a * b / gcd<T>(a, b);
}
template <class T>
inline T power(T b, T p) {
  long long ans = 1;
  while (p--) ans *= b;
  return ans;
}
long long binomialCoeff(long long n, long long r) {
  if (r > n) return 0;
  long long m = 998244353;
  long long inv[r + 1];
  memset(inv, 0, sizeof(inv));
  inv[0] = 1;
  if (r + 1 >= 2) inv[1] = 1;
  for (int i = 2; i <= r; i++) {
    inv[i] = m - (m / i) * inv[m % i] % m;
  }
  long long ans = 1;
  for (int i = 2; i <= r; i++) {
    ans = ((ans % m) * (inv[i] % m)) % m;
  }
  for (int i = n; i >= (n - r + 1); i--) {
    ans = ((ans % m) * (i % m)) % m;
  }
  return ans;
}
long long extgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long x1, y1;
  long long d = extgcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return d;
}
bool find_any_solution(long long a, long long b, long long c, long long &x0,
                       long long &y0, long long &g) {
  g = extgcd(abs(a), abs(b), x0, y0);
  if (c % g) {
    return false;
  }
  x0 *= c / g;
  y0 *= c / g;
  if (a < 0) x0 = -x0;
  if (b < 0) y0 = -y0;
  return true;
}
long long mul(long long a, long long b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
long long add(long long a, long long b) {
  return (a % 1000000007 + b % 1000000007) % 1000000007;
}
long long powm(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
long long divide(long long a, long long b) {
  return (a % 1000000007) * powm(b, 1000000007 - 2) % 1000000007;
}
bool vectorSortedOrNot(vector<int> v, int n) {
  if (n == 1 || n == 0) return 1;
  if (v[n - 1] < v[n - 2]) return 0;
  return vectorSortedOrNot(v, n - 1);
}
int sumDigits(int num) { return num == 0 ? 0 : num % 10 + sumDigits(num / 10); }
void Divisors(long long n) {
  for (long long i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
    }
  }
}
long long solve(long long mid, long long k) {
  long long sum = 0;
  long long z = 1;
  while (mid / z > 0) {
    sum += mid / z;
    z *= k;
  }
  return sum;
}
pair<int, int> solve(string s) {
  int a(0), b(0);
  int n = s.size();
  for (int i = 0; i < n - 1; i++) {
    if (s[i] == 'a' and s[i + 1] == 'b')
      a++;
    else if (s[i] == 'b' and s[i + 1] == 'a')
      b++;
  }
  pair<int, int> ab;
  ab.first = a;
  ab.second = b;
  return ab;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed << setprecision(0);
  long long test = 1;
  cin >> test;
  while (test--) {
    long long n, k;
    cin >> n >> k;
    vector<int> v(n);
    long long ans = 0;
    k += 1;
    for (long long i = 0; i < n; i++) cin >> v[i];
    for (long long i = 0; i < n - 1; i++) {
      if (k > 0) {
        long long x = powm(10, v[i + 1] - v[i]) - 1;
        long long val = min(x, k);
        ans += powm(10, v[i]) * val;
        k -= val;
      }
    }
    ans += k * powm(10, v[n - 1]);
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to 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;
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
inline bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.second == b.second) return a.first < b.first;
  return a.second > b.second;
}
int drx[8] = {-2, -2, -1, -1, 1, 1, 2, 2};
int dcy[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
int dirx[4] = {-1, 0, 1, 0};
int diry[4] = {0, -1, 0, 1};
long long binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long InverseMod(long long base, long long pow) {
  if (pow == 0) return 1;
  long long ans = InverseMod(base, pow / 2);
  ans = (ans * ans) % 1000000007;
  if (pow & 1) {
    return (ans * base) % 1000000007;
  } else {
    return ans;
  }
}
bool isprime(long long num) {
  if (num < 2) return false;
  for (long long i = 2; i * i <= num; i++) {
    if (num % i == 0) return false;
  }
  return true;
}
long long EularPHI(long long num) {
  double ans = num;
  for (long long i = 2; i * i <= num; i++) {
    if (num % i == 0) {
      while (num % i == 0) {
        num /= i;
      }
      ans *= (1.0 - (1.0 / (double)i));
    }
  }
  if (num > 1) ans *= (1.0 - (1.0 / (double)num));
  return (long long)ans;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
  return a * b / gcd<T>(a, b);
}
template <class T>
inline T power(T b, T p) {
  long long ans = 1;
  while (p--) ans *= b;
  return ans;
}
long long binomialCoeff(long long n, long long r) {
  if (r > n) return 0;
  long long m = 998244353;
  long long inv[r + 1];
  memset(inv, 0, sizeof(inv));
  inv[0] = 1;
  if (r + 1 >= 2) inv[1] = 1;
  for (int i = 2; i <= r; i++) {
    inv[i] = m - (m / i) * inv[m % i] % m;
  }
  long long ans = 1;
  for (int i = 2; i <= r; i++) {
    ans = ((ans % m) * (inv[i] % m)) % m;
  }
  for (int i = n; i >= (n - r + 1); i--) {
    ans = ((ans % m) * (i % m)) % m;
  }
  return ans;
}
long long extgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long x1, y1;
  long long d = extgcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return d;
}
bool find_any_solution(long long a, long long b, long long c, long long &x0,
                       long long &y0, long long &g) {
  g = extgcd(abs(a), abs(b), x0, y0);
  if (c % g) {
    return false;
  }
  x0 *= c / g;
  y0 *= c / g;
  if (a < 0) x0 = -x0;
  if (b < 0) y0 = -y0;
  return true;
}
long long mul(long long a, long long b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
long long add(long long a, long long b) {
  return (a % 1000000007 + b % 1000000007) % 1000000007;
}
long long powm(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
long long divide(long long a, long long b) {
  return (a % 1000000007) * powm(b, 1000000007 - 2) % 1000000007;
}
bool vectorSortedOrNot(vector<int> v, int n) {
  if (n == 1 || n == 0) return 1;
  if (v[n - 1] < v[n - 2]) return 0;
  return vectorSortedOrNot(v, n - 1);
}
int sumDigits(int num) { return num == 0 ? 0 : num % 10 + sumDigits(num / 10); }
void Divisors(long long n) {
  for (long long i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
    }
  }
}
long long solve(long long mid, long long k) {
  long long sum = 0;
  long long z = 1;
  while (mid / z > 0) {
    sum += mid / z;
    z *= k;
  }
  return sum;
}
pair<int, int> solve(string s) {
  int a(0), b(0);
  int n = s.size();
  for (int i = 0; i < n - 1; i++) {
    if (s[i] == 'a' and s[i + 1] == 'b')
      a++;
    else if (s[i] == 'b' and s[i + 1] == 'a')
      b++;
  }
  pair<int, int> ab;
  ab.first = a;
  ab.second = b;
  return ab;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed << setprecision(0);
  long long test = 1;
  cin >> test;
  while (test--) {
    long long n, k;
    cin >> n >> k;
    vector<int> v(n);
    long long ans = 0;
    k += 1;
    for (long long i = 0; i < n; i++) cin >> v[i];
    for (long long i = 0; i < n - 1; i++) {
      if (k > 0) {
        long long x = powm(10, v[i + 1] - v[i]) - 1;
        long long val = min(x, k);
        ans += powm(10, v[i]) * val;
        k -= val;
      }
    }
    ans += k * powm(10, v[n - 1]);
    cout << ans << endl;
  }
  return 0;
}

```