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;
int n, m, k, a, b;
vector<int> v;
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> a;
    v.push_back(a);
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < v.size(); i++) {
    b = b + (v[v.size() - 1] - v[i]);
  }
  if (b >= m) {
    cout << v[v.size() - 1];
  } else {
    if ((m - b) % v.size() == 0) {
      cout << (m - b) / v.size() + v[v.size() - 1];
    } else {
      cout << (m - b) / v.size() + 1 + v[v.size() - 1];
    }
  }
  cout << " " << v[v.size() - 1] + m;
  return 0;
}
