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

public class CF1417E extends PrintWriter {
	CF1417E() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1417E o = new CF1417E(); o.main(); o.flush();
	}

	static final int B = 30;
	long[][] inv = new long[B][2];
	int[] aa;
	void solve(int[] ii, int[] tt, int l, int r, int h) {
		if (h == -1)
			return;
		int b = 1 << h, n0 = 0, n1 = 0, i0 = l, i1 = r - 1;
		for (int i = l; i < r; i++) {
			int i_ = ii[i];
			if ((aa[i_] & b) == 0) {
				inv[h][0] += n1; n0++;
				tt[i0++] = i_;
			} else {
				inv[h][1] += n0; n1++;
				tt[i1--] = i_;
			}
		}
		if (n0 > 0)
			solve(tt, ii, l, l + n0, h - 1);
		if (n1 > 0) {
			for (int i = l + n0, j = r - 1; i < j; i++, j--) {
				int tmp = tt[i]; tt[i] = tt[j]; tt[j] = tmp;
			}
			solve(tt, ii, l + n0, r, h - 1);
		}
	}
	void main() {
		int n = sc.nextInt();
		aa = new int[n];
		int[] ii = new int[n], tt = new int[n];
		for (int i = 0; i < n; i++) {
			aa[i] = sc.nextInt();
			ii[i] = i;
		}
		solve(ii, tt, 0, n, B - 1);
		long k = 0; int x = 0;
		for (int h = 0; h < B; h++)
			if (inv[h][0] <= inv[h][1])
				k += inv[h][0];
			else {
				k += inv[h][1];
				x |= 1 << h;
			}
		println(k + " " + x);
	}
}
