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;
template <typename T>
void read(T &x) {
  x = 0;
  char c = getchar();
  int sgn = 1;
  while (c < '0' || c > '9') {
    if (c == '-') sgn = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  x *= sgn;
}
template <typename T>
void out(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x >= 10) out(x / 10);
  putchar(x % 10 + '0');
}
int a[1005];
int main() {
  int n;
  read(n);
  int m;
  read(m);
  for (int i = 0; i < n; i++) read(a[i]);
  sort(a, a + n);
  int total = 0;
  for (int i = 0; i < n - 1; i++) total += a[n - 1] - a[i];
  if (m <= total) {
    cout << a[n - 1] << " " << a[n - 1] + m << endl;
  } else {
    if ((m - total) % n == 0)
      cout << a[n - 1] + (m - total) / n << " " << a[n - 1] + m << endl;
    else
      cout << a[n - 1] + (m - total) / n + 1 << " " << a[n - 1] + m << endl;
  }
  return 0;
}
