Problem:
Score : 1000 points

Problem Statement

There is a sequence of length N: a = (a_1, a_2, ..., a_N).
Here, each a_i is a non-negative integer.

Snuke can repeatedly perform the following operation:

Let the XOR of all the elements in a be x. Select an integer i (1 ≤ i ≤ N) and replace a_i with x.

Snuke's objective is to match a with another sequence b = (b_1, b_2, ..., b_N).
Here, each b_i is a non-negative integer.

Determine whether the objective is achievable, and find the minimum necessary number of operations if the answer is positive.

Constraints

2 ≤ N ≤ 10^5

a_i and b_i are integers.

0 ≤ a_i, b_i < 2^{30}

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N
b_1 b_2 ... b_N

Output

If the objective is achievable, print the minimum necessary number of operations.
Otherwise, print -1 instead.

Sample Input 1

3
0 1 2
3 1 0

Sample Output 1

2

At first, the XOR of all the elements of a is 3.
If we replace a_1 with 3, a becomes (3, 1, 2).

Now, the XOR of all the elements of a is 0.
If we replace a_3 with 0, a becomes (3, 1, 0), which matches b.

Sample Input 2

3
0 1 2
0 1 2

Sample Output 2

0

Sample Input 3

2
1 1
0 0

Sample Output 3

-1

Sample Input 4

4
0 1 2 3
1 0 3 2

Sample Output 4

5

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
#include<map>
#include<set>
#define ll long long
#define M 200010
#define mmp make_pair
using namespace std;
int read() {
	int nm = 0, f = 1;
	char c = getchar();
	for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
	return nm * f;
}
ll a[M], b[M], n, ans, tm, now, tp, an[M], bn[M], sta[M], tot;
multiset<ll> st;
multiset<ll>:: iterator it;
ll bude = 0;
map<ll, int> mp;
ll tmp;
int father[M];
int find(int x) {
	return father[x] == x ? x : father[x] = find(father[x]);
}
int main() {
//	freopen("replace.in", "r", stdin);
	//freopen("replace.out", "w", stdout);
	n = read();
	for(int i = 1; i <= n; i++) a[i] = read(), now ^= a[i];
	for(int i = 1; i <= n; i++) b[i] = read(), st.insert(b[i]);
	for(int i = 1; i <= n; i++) {
		it = st.find(a[i]);
		if(it == st.end()) tm++, bude = a[i];
		else st.erase(it);
		if(a[i] == b[i]) continue;
		ans++;
	}
	for(int i = 1; i <= n; i++) {
		if(a[i] != b[i]) break;
		if(i == n) {
			cout << "0\n";
			return 0;
		}
	}
	if(st.find(now) == st.end()) tm++, bude = now;
	else st.erase(st.find(now));
	if(tm >= 2 || st.size()) {
		cout << "-1" << "\n";
		return 0;
	}
	tp = 0, tot = 0;
	for(int i = 1; i <= n; i++) {
		if(a[i] == b[i]) continue;
		sta[++tot] = a[i];
		an[++tp] = a[i];
		bn[tp] = b[i];
	}
	sta[0] = -1;
	sort(sta + 1, sta + tot + 1);
	int cnt = 0;
	for(int i = 1; i <= tot; i++) {
		if(sta[i] == sta[i - 1]) continue;
		mp[sta[i]] = ++cnt;
	}
	for(int i = 1; i <= cnt + 1; i++) father[i] = i;
	for(int i = 1; i <= tp; i++) {
		int vi = mp[an[i]], vj = mp[bn[i]];
		vi = find(vi), vj = find(vj);
		//	cout << vi << " " << vj << "\n";
		if(vi == vj) continue;
		father[vi] = vj;
	}
	ans--;
	for(int i = 1; i <= n; i++) {
		if(a[i] == b[i]) continue;
		if(b[i] == now) break;
		if(i == n) {
			ans++;
		}
	}
	for(int i = 1; i <= cnt + 1; i++) if(father[i] == i) ans++;
	cout << ans - 1<< "\n";
	return 0;
}
/*
5
1 2 3 0 5
1 2 3 5 5

1
2
2
*/