技術標籤：Codeforces

Today, as a friendship gift, Bakry gave Badawy n n n integers a 1 , a 2 , … , a n a_1,a_2,…,a_n a1​,a2​,…,an​ and challenged him to choose an integer X X X such that the value m a x 1 ≤ i ≤ n ( a i ⊕ X ) \underset{1≤i≤n}{max}(a_i⊕X) 1≤i≤nmax​(ai​⊕X) is minimum possible, where ⊕ ⊕

As always, Badawy is too lazy, so you decided to help him and find the minimum possible value of m a x 1 ≤ i ≤ n ( a i ⊕ X ) \underset{1≤i≤n}{max}(a_i⊕X) 1≤i≤nmax​(ai​⊕X)

Input

The first line contains integer n ( 1 ≤ n ≤ 1 0 5 ) n (1≤n≤10^5) n(1≤n≤105).

The second line contains n n

Output

Print one integer — the minimum possible value of m a x 1 ≤ i ≤ n ( a i ⊕ X ) \underset{1≤i≤n}{max}(a_i⊕X) 1≤i≤nmax​(ai​⊕X)

Examples

input

3
1 2 3

output

2

input

2
1 5

output

4
 

Note

In the first sample, we can choose X = 3 X=3 X=3.

In the second sample, we can choose X = 5 X=5 X=5.

#include <bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
int n;
int trie[maxn*32][3],cnt=1,ans;
void insert(int x) {
    int p = 1;
    for (int i = 30; i >= 0; i--) {
        int now = (x >> i) & 1;
        if (!trie[p][now]) {
            trie[p][now] = ++cnt;
        }
        p = trie[p][now];
    }
}
int search(int cnt,int now) {
    if (cnt == -1) return 0;
    if (trie[now][0] == 0) {
        search(cnt - 1, trie[now][1]);
    } else if (trie[now][1] == 0) {
        search(cnt - 1, trie[now][0]);
    } else {
        return (1 << cnt) + min(search(cnt - 1, trie[now][0]), search(cnt - 1, trie[now][1]));
    }
}
int main() {
    scanf("%d", &n);
    for (int i = 1, x; i <= n; i++) {
        scanf("%d", &x);
        insert(x);
    }

    printf("%d\n", search(30, 1));
}