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.
import java.util.*;
import java.io.*;

public class a 
{
    static boolean multiple_testcases = false;
        
    static void solve() {
        int n = in.nextInt(), m = in.nextInt();
        int ans=0, max=0, min= INF, sum = m;
        for(int i=0;i<n;i++) {
            int ele = in.nextInt();
            sum += ele;
            if(ele > max) max = ele;
            if(ele < min) min = ele;
        }

        int low = min, high = max+m;
        while(low <= high) {
            int mid = low + (high - low)/2;
            if(n*mid >= sum) {
                ans = mid;
                high = mid -1;
            } else low = mid +1;
        }

        System.out.println(Math.max(ans,max)+" "+(max+m));
    }

    static int INF = Integer.MAX_VALUE;
    static FastReader in = new FastReader();
    public static void main(String[] args) { 
        if(multiple_testcases) {
            int t = in.nextInt();
            while(t-- >0)  solve();
        } else solve();
    }

    static class FastReader { 
        BufferedReader br; StringTokenizer st; 
        public FastReader() { br = new BufferedReader(new InputStreamReader(System.in)); } 
      
        String next() { while (st == null || !st.hasMoreElements()) { 
            try{ st = new StringTokenizer(br.readLine()); } catch (IOException  e) { e.printStackTrace(); } }
            return st.nextToken(); } 
        int nextInt() { return Integer.parseInt(next()); } 
        long nextLong() { return Long.parseLong(next()); } 
        double nextDouble() { return Double.parseDouble(next()); } 
        char nextChar() { return next().charAt(0); }
        String nextLine() { String str = ""; try { str = br.readLine(); } catch (IOException e) { e.printStackTrace(); } 
        return str; }
     
        int[] intArray(int n) {
            int[] arr = new int[n];
            for(int i=0;i<n;i++) 
                arr[i] = in.nextInt();
            return arr;    
        } 
        int[] intArray1(int n) {
            int[] arr = new int[n+1];
            for(int i=1;i<=n;i++) 
                arr[i] = in.nextInt();
            return arr;    
        } 
    } 
}       