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;
const int MOD = 1e9 + 7;
int main() {
  std ::ios_base ::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  int mx_ans = *max_element(a.begin(), a.end());
  int mn_ans = mx_ans;
  mx_ans += m;
  while (1) {
    if (m == 0) {
      cout << mn_ans << " " << mx_ans << endl;
      break;
    }
    int mx = max_element(a.begin(), a.end()) - a.begin();
    int mn = min_element(a.begin(), a.end()) - a.begin();
    if (a[mx] == a[mn]) {
      mn_ans = a[mn] + ceil((m * 1.0) / n);
      m = 0;
      continue;
    }
    if (m >= a[mx] - a[mn]) {
      m -= (a[mx] - a[mn]);
      a[mn] = a[mx];
    } else {
      m = 0;
      mn_ans = a[mx];
    }
  }
}
