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>
template <typename T>
std::vector<T> readVectorSz(int sz) {
  std::vector<T> res;
  res.reserve(sz);
  for (int i = 0; i < sz; ++i) {
    T tmp;
    std::cin >> tmp;
    res.push_back(std::move(tmp));
  }
  return res;
}
template <typename T>
std::vector<T> readVector() {
  int sz;
  std::cin >> sz;
  return readVectorSz<T>(sz);
}
std::vector<std::vector<int>> readGraphDir(int vertices, int edges) {
  auto g = std::vector<std::vector<int>>(vertices, std::vector<int>());
  for (int i = 0; i < edges; ++i) {
    int u, v;
    std::cin >> u >> v;
    --u;
    --v;
    g[u].push_back(v);
  }
  return g;
}
std::vector<std::vector<int>> readGraphUn(int vertices, int edges) {
  auto g = std::vector<std::vector<int>>(vertices, std::vector<int>());
  for (int i = 0; i < edges; ++i) {
    int u, v;
    std::cin >> u >> v;
    --u;
    --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  return g;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(NULL);
  int n, m;
  std::cin >> n >> m;
  auto vec = readVectorSz<int>(n);
  int max = *std::max_element(vec.cbegin(), vec.cend());
  int rem = 0;
  for (auto i : vec) rem += max - i;
  int m_new = std::max(0, m - rem);
  std::cout << max + ((m_new + n - 1) / n) << ' ' << max + m << '\n';
}
