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;
bool sortbysecdesc(const pair<int, int> &a, const pair<int, int> &b) {
  return a.second > b.second;
}
bool sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first > b.first);
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
int findGCD(long long a[], long long n) {
  int result = a[0];
  for (int i = 1; i < n; i++) {
    result = gcd(a[i], result);
    if (result == 1) {
      return 1;
    }
  }
  return result;
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long n, m, p, ans2 = INT_MIN;
  cin >> n >> m;
  long long a[n];
  for (int i = 0; i < int(n); i++) cin >> a[i];
  p = *max_element(a, a + n) + m;
  sort(a, a + n);
  ans2 = p - m;
  while (m) {
    long long mi = 0;
    for (int i = 1; i < n; i++) {
      if (a[i] < a[mi]) mi = i;
    }
    m--;
    a[mi]++;
    ans2 = max(ans2, a[mi]);
  }
  cout << ans2 << " " << p << endl;
  return 0;
}
