CodeForces

You are given an array a1,a2,…,an

of integer numbers.

Your task is to divide the array into the maximum number of segments in such a way that:

• each element is contained in exactly one segment;
• each segment contains at least one element;
• there doesn't exist a non-empty subset of segments such that bitwise XOR of the numbers from them is equal to 0

• .

Print the maximum number of segments the array can be divided into. Print -1 if no suitable division exists.

Input

The first line contains a single integer n

(1≤n≤2⋅105

) — the size of the array.

The second line contains n

integers a1,a2,…,an (0≤ai≤109

).

Output

Print the maximum number of segments the array can be divided into while following the given constraints. Print -1 if no suitable division exists.

Examples

Input

4
5 5 7 2

Output

2

Input

3
1 2 3

Output

-1

Input

3
3 1 10

Output

3

题意：给你一些树，让你给树分组，然后把每一个分组看成元素，满足没有元素集合的异或和位0。

思路：首先知道一点，我们得到线性基的时候，当1的个数num（基底个数）不等于元素个数时，表示可以拼凑出0 ；那么，如果全部数的异或为0，那么无解；否则，我们现在得到了线性基，最多划分为几个集合呢，答案就是num。 因为这num个位置的最高位1的位置不同，他们是线性无关的。

（给出N个数,要从中选出一个最大的子集,使得子集中的任意个元素异或值不为0，答案也是基底

#include<bits/stdc++.h>
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define rep2(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
int p[66],ans;
int main()
{
    int N,sum=0,num=0,x;
    scanf("%d",&N);
    rep(i,1,N) {
        scanf("%d",&x); sum^=x;
        rep2(j,30,0){
            if(x&(1LL<<j)){
                if(p[j]) x^=p[j];
                else { p[j]=x;break;}
            }
        }
    }
    if(sum==0) return puts("-1"),0;
    rep(i,0,30) if(p[i]){
        p[num++]=p[i];
        rep(j,i+1,30) if((p[j]>>i)&1) p[j]^=p[i];
    }
    printf("%d
",num);
    return 0;
}