There are n benches in the Berland Central park. It is known that a_i people are currently sitting on the i-th bench. Another m people are coming to the park and each of them is going to have a seat on some bench out of n available.

Let k be the maximum number of people sitting on one bench after additional m people came to the park. Calculate the minimum possible k and the maximum possible k.

Nobody leaves the taken seat during the whole process.

Input

The first line contains a single integer n (1 ≤ n ≤ 100) — the number of benches in the park.

The second line contains a single integer m (1 ≤ m ≤ 10 000) — the number of people additionally coming to the park.

Each of the next n lines contains a single integer a_i (1 ≤ a_i ≤ 100) — the initial number of people on the i-th bench.

Output

Print the minimum possible k and the maximum possible k, where k is the maximum number of people sitting on one bench after additional m people came to the park.

Examples

Input

4
6
1
1
1
1


Output

3 7


Input

1
10
5


Output

15 15


Input

3
6
1
6
5


Output

6 12


Input

3
7
1
6
5


Output

7 13

Note

In the first example, each of four benches is occupied by a single person. The minimum k is 3. For example, it is possible to achieve if two newcomers occupy the first bench, one occupies the second bench, one occupies the third bench, and two remaining — the fourth bench. The maximum k is 7. That requires all six new people to occupy the same bench.

The second example has its minimum k equal to 15 and maximum k equal to 15, as there is just a single bench in the park and all 10 people will occupy it.
#include <bits/stdc++.h>
using namespace std;
inline int add(int a, int b) {
  a += b;
  if (a >= 1000000007) a -= 1000000007;
  return a;
}
inline int sub(int a, int b) {
  a -= b;
  if (a < 0) a += 1000000007;
  return a;
}
inline int mul(int a, int b) { return (a * 1ll * b) % 1000000007; }
inline int power(int a, int b) {
  int rt = 1;
  while (b > 0) {
    if (b & 1) rt = mul(rt, a);
    a = mul(a, a);
    b >>= 1;
  }
  return rt;
}
inline int inv(int a) { return power(a, 1000000007 - 2); }
const string yes = "YES";
const string no = "NO";
void _print(long long t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(unsigned long long t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T>
void _print(unordered_set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T, class V>
void _print(unordered_map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T>
void _print(deque<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(unordered_set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(unordered_map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(deque<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << " ]";
}
void printGraph(vector<int> adj[], int V) {
  for (int v = 0; v < V; ++v) {
    cerr << "\n Adjacency list of vertex " << v << "\n head ";
    for (auto x : adj[v]) cerr << "-> " << x;
    cerr << "\n";
  }
}
void solve() {
  long long n, m;
  cin >> n >> m;
  vector<long long> arr(n);
  priority_queue<long long, vector<long long>, greater<long long>> pq;
  for (auto &it : arr) {
    cin >> it;
    pq.push(it);
  }
  long long temp = m;
  long long maxi = *max_element((arr).begin(), (arr).end());
  while (m--) {
    long long curr = pq.top();
    pq.pop();
    pq.push(curr + 1);
  }
  while (pq.size() != 1) pq.pop();
  while (pq.size()) {
    cerr << pq.top() << " ";
    pq.pop();
  }
  cout << pq.top() << " " << temp + maxi << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
}
