You are given an array a consisting of n non-negative integers. You have to choose a non-negative integer x and form a new array b of size n according to the following rule: for all i from 1 to n, b_i = a_i ⊕ x (⊕ denotes the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)).

An inversion in the b array is a pair of integers i and j such that 1 ≤ i < j ≤ n and b_i > b_j.

You should choose x in such a way that the number of inversions in b is minimized. If there are several options for x — output the smallest one.

Input

First line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of elements in a.

Second line contains n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.

Output

Output two integers: the minimum possible number of inversions in b, and the minimum possible value of x, which achieves those number of inversions.

Examples

Input


4
0 1 3 2


Output


1 0


Input


9
10 7 9 10 7 5 5 3 5


Output


4 14


Input


3
8 10 3


Output


0 8

Note

In the first sample it is optimal to leave the array as it is by choosing x = 0.

In the second sample the selection of x = 14 results in b: [4, 9, 7, 4, 9, 11, 11, 13, 11]. It has 4 inversions:

  * i = 2, j = 3; 
  * i = 2, j = 4; 
  * i = 3, j = 4; 
  * i = 8, j = 9. 



In the third sample the selection of x = 8 results in b: [0, 2, 11]. It has no inversions.
import java.util.*;
import java.lang.*;
import java.io.*;

public class Main
{
	PrintWriter out = new PrintWriter(System.out);
	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer tok = new StringTokenizer("");
    String next() throws IOException {
        if (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); }
        return tok.nextToken();
    }
    int ni() throws IOException { return Integer.parseInt(next()); }
    long nl() throws IOException { return Long.parseLong(next()); }
    
    long mod=1000000007;
    int n;
    int[]A;
    long[]leave,swap;
    
    void solve() throws IOException {
        n=ni();
        A=new int[n+1];
        leave=new long[32];
        swap=new long[31];
        ArrayList<Integer>B=new ArrayList();
        for (int i=1;i<=n;i++) { B.add(i); A[i]=ni(); }
        
        process(B,30);
        
        int m=1;
        int ans=0;
        long num=0;
        for (int i=0;i<31;i++) {
            if (swap[i]<leave[i]) { ans|=m; num+=swap[i]; }
            else num+=leave[i];
            m<<=1;
        }
        
        out.println(num+" "+ans);
        out.flush();
    }
    
    void process(ArrayList<Integer> B, int lev) {
        if (lev<0 || B.size()<2) return;
        
        ArrayList<Integer>X=new ArrayList();
        ArrayList<Integer>Y=new ArrayList();
        int m=1<<lev;
        
        long leavepre=0;
        long swappre=0;
        long leaveinv=0;
        long swapinv=0;
        
        for (Integer v:B) {
            if ((A[v]&m)>0) {
                X.add(v);
                leavepre++;
                swapinv+=swappre;
            }
            else {
                Y.add(v);
                leaveinv+=leavepre;
                swappre++;
            }
        }
        
        swap[lev]+=swapinv;
        leave[lev]+=leaveinv;
        
        process(X,lev-1);
        process(Y,lev-1);
    }
    
    int gcd(int a,int b) { return(b==0?a:gcd(b,a%b)); }
    long gcd(long a,long b) { return(b==0?a:gcd(b,a%b)); }
    long mp(long a,long p) { long r=1; while(p>0) { if ((p&1)==1) r=(r*a)%mod; p<<=1; a=(a*a)%mod; } return r; }
    
    public static void main(String[] args) throws IOException {
        new Main().solve();
    }
}