Alice has a very important message M consisting of some non-negative integers that she wants to keep secret from Eve. Alice knows that the only theoretically secure cipher is one-time pad. Alice generates a random key K of the length equal to the message’s length. Alice computes the bitwise xor of each element of the message and the key (, where denotes the bitwise XOR operation) and stores this encrypted message A. Alice is smart. Be like Alice.
For example, Alice may have wanted to store a message M = (0, 15, 9, 18). She generated a key K = (16, 7, 6, 3). The encrypted message is thus A = (16, 8, 15, 17).
Alice realised that she cannot store the key with the encrypted message. Alice sent her key K to Bob and deleted her own copy. Alice is smart. Really, be like Alice.
Bob realised that the encrypted message is only secure as long as the key is secret. Bob thus randomly permuted the key before storing it. Bob thinks that this way, even if Eve gets both the encrypted message and the key, she will not be able to read the message. Bob is not smart. Don’t be like Bob.
In the above example, Bob may have, for instance, selected a permutation (3, 4, 1, 2) and stored the permuted key P = (6, 3, 16, 7).
One year has passed and Alice wants to decrypt her message. Only now Bob has realised that this is impossible. As he has permuted the key randomly, the message is lost forever. Did we mention that Bob isn’t smart?
Bob wants to salvage at least some information from the message. Since he is not so smart, he asks for your help. You know the encrypted message A and the permuted key P. What is the lexicographically smallest message that could have resulted in the given encrypted text?
More precisely, for given A and P, find the lexicographically smallest message O, for which there exists a permutation π such that for every i.
Note that the sequence S is lexicographically smaller than the sequence T, if there is an index i such that Si < Ti and for all j < i the condition Sj = Tj holds.
Input
The first line contains a single integer N (1 ≤ N ≤ 300000), the length of the message.
The second line contains N integers A1, A2, …, AN (0 ≤ Ai < 230) representing the encrypted message.
The third line contains N integers P1, P2, …, PN (0 ≤ Pi < 230) representing the permuted encryption key.
Output
Output a single line with N integers, the lexicographically smallest possible message O. Note that all its elements should be non-negative.
Examples
Input
3
8 4 13
17 2 7
Output
10 3 28
Input
5
12 7 87 22 11
18 39 9 12 16
Output
0 14 69 6 44
Input
10
331415699 278745619 998190004 423175621 42983144 166555524 843586353 802130100 337889448 685310951
226011312 266003835 342809544 504667531 529814910 684873393 817026985 844010788 993949858 1031395667
Output
128965467 243912600 4281110 112029883 223689619 76924724 429589 119397893 613490433 362863284
异或之后数值最小，那么我们先建立01树，匹配的时候尽量找相同的路径走自然是最小，如果没有节点了才走相反的节点；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<ctime>
typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 500005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const long long int mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair<int,int>
#define eps 1e-5
#define pll pair<ll,ll>
ll quickpow(ll a, ll b) {
ll ans = 1;
while (b > 0) {
if (b % 2)ans = ans * a;
b = b / 2;
a = a * a;
}
return ans;
}
int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a%b);
}
int n, a[maxn], p[maxn], trisize = 1;
struct Trie {
int cnt, zero, one;
}trie[maxn*35];
void Insert(int x) {
int cur = 1;
trie[1].cnt++;
for (int i = 29; i >= 0; i--) {
if ((x >> i) & 1) {
if (!trie[cur].one)trie[cur].one = ++trisize;
cur = trie[cur].one;
}
else {
if (!trie[cur].zero)trie[cur].zero = ++trisize;
cur = trie[cur].zero;
}
trie[cur].cnt++;
}
}
int query(int x) {
int cur = 1, nxt;
int ret = 0;
trie[1].cnt--;
for (int i = 29; i >= 0; i--) {
if ((x >> i) & 1) {
nxt = trie[cur].one;
if (trie[nxt].cnt == 0)nxt = trie[cur].zero;
}
else {
nxt = trie[cur].zero;
if (trie[nxt].cnt == 0)nxt = trie[cur].one;
}
ret <<= 1;
if (nxt == trie[cur].one)ret++;
cur = nxt;
trie[cur].cnt--;
}
return ret;
}
int main()
{
ios::sync_with_stdio(false);
cin >> n;
for (int i = 0; i < n; i++)cin >> a[i];
for (int i = 0; i < n; i++) {
cin >> p[i];
Insert(p[i]);
}
for (int i = 0; i < n; i++) {
printf("%d ", a[i] ^ query(a[i]));
//cout << (a[i] ^ query(a[i])) << ' ';
}
}