#include <bits/stdc++.h>
using namespace std;
void In() { freopen("in.in", "r", stdin); }
void Out() { freopen("out.out", "w", stdout); }
const int N = 2e5 + 10;
const int M = 3e5 + 10;
const int Mbit = 1e6 + 10;
const int inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
int n;
long long suma, sumd;
long long a[N], b[N], c[N], d[N], base[40];
int B[N][40], C[N][40], k[N];
int main() {
  base[0] = 1;
  for (int i = 1; i <= 32; i++) base[i] = base[i - 1] * 2;
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) scanf("%lld", &b[i]);
    for (int i = 1; i <= n; i++) scanf("%lld", &c[i]);
    sumd = 0;
    for (int i = 1; i <= n; i++) d[i] = b[i] + c[i], sumd += d[i];
    suma = sumd / 2 / n;
    for (int i = 1; i <= n; i++) a[i] = (d[i] - suma) / n;
    for (int i = 0; i <= 32; i++) {
      long long x = 1ll << i;
      for (int j = 1; j <= n; j++)
        if (a[j] & x) k[i]++;
    }
    for (int i = 0; i <= 32; i++) {
      long long x = 1ll << i;
      for (int j = 1; j <= n; j++)
        if (a[j] & x) {
          B[j][i] = k[i];
          C[j][i] = n;
        } else {
          B[j][i] = 0;
          C[j][i] = k[i];
        }
    }
    bool flag = 1;
    for (int i = 1; i <= n; i++) {
      long long res = 0;
      for (int j = 0; j <= 32; j++) res = res + B[i][j] * base[j];
      if (res != b[i]) {
        flag = 0;
        break;
      }
      res = 0;
      for (int j = 0; j <= 32; j++) res = res + C[i][j] * base[j];
      if (res != c[i]) {
        flag = 0;
        break;
      }
    }
    if (!flag)
      puts("-1");
    else {
      for (int i = 1; i <= n; i++)
        printf("%lld%c", a[i], (i == n) ? '\n' : ' ');
    }
  }
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

Anton goes to school, his favorite lessons are arraystudying. He usually solves all the tasks pretty fast, but this time the teacher gave him a complicated one: given two arrays b and c of length n, find array a, such that:

<image>

where a and b means bitwise AND, while a or b means bitwise OR.

Usually Anton is good in arraystudying, but this problem is too hard, so Anton asks you to help.

Input

The first line of the input contains a single integers n (1 ≤ n ≤ 200 000) — the size of arrays b and c.

The second line contains n integers bi (0 ≤ bi ≤ 109) — elements of the array b.

Third line contains n integers ci (0 ≤ ci ≤ 109) — elements of the array c.

Output

If there is no solution, print  - 1.

Otherwise, the only line of the output should contain n non-negative integers ai — elements of the array a. If there are multiple possible solutions, you may print any of them.

Examples

Input

4
6 8 4 4
16 22 10 10


Output

3 5 1 1 


Input

5
8 25 14 7 16
19 6 9 4 25


Output

-1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void In() { freopen("in.in", "r", stdin); }
void Out() { freopen("out.out", "w", stdout); }
const int N = 2e5 + 10;
const int M = 3e5 + 10;
const int Mbit = 1e6 + 10;
const int inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
int n;
long long suma, sumd;
long long a[N], b[N], c[N], d[N], base[40];
int B[N][40], C[N][40], k[N];
int main() {
  base[0] = 1;
  for (int i = 1; i <= 32; i++) base[i] = base[i - 1] * 2;
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; i++) scanf("%lld", &b[i]);
    for (int i = 1; i <= n; i++) scanf("%lld", &c[i]);
    sumd = 0;
    for (int i = 1; i <= n; i++) d[i] = b[i] + c[i], sumd += d[i];
    suma = sumd / 2 / n;
    for (int i = 1; i <= n; i++) a[i] = (d[i] - suma) / n;
    for (int i = 0; i <= 32; i++) {
      long long x = 1ll << i;
      for (int j = 1; j <= n; j++)
        if (a[j] & x) k[i]++;
    }
    for (int i = 0; i <= 32; i++) {
      long long x = 1ll << i;
      for (int j = 1; j <= n; j++)
        if (a[j] & x) {
          B[j][i] = k[i];
          C[j][i] = n;
        } else {
          B[j][i] = 0;
          C[j][i] = k[i];
        }
    }
    bool flag = 1;
    for (int i = 1; i <= n; i++) {
      long long res = 0;
      for (int j = 0; j <= 32; j++) res = res + B[i][j] * base[j];
      if (res != b[i]) {
        flag = 0;
        break;
      }
      res = 0;
      for (int j = 0; j <= 32; j++) res = res + C[i][j] * base[j];
      if (res != c[i]) {
        flag = 0;
        break;
      }
    }
    if (!flag)
      puts("-1");
    else {
      for (int i = 1; i <= n; i++)
        printf("%lld%c", a[i], (i == n) ? '\n' : ' ');
    }
  }
  return 0;
}

```