#include <bits/stdc++.h>
using namespace std;
const int N = 10005;
int n, a[N], b[N];
vector<pair<int, int> > qa, qb;
void add(int *a, int i, int j, vector<pair<int, int> > &q) {
  a[i] ^= a[j];
  q.push_back(pair<int, int>(i, j));
}
void work(int *a, vector<pair<int, int> > &q) {
  for (int i = 30, j = 1, k; i >= 0; i--) {
    for (k = j; k <= n && !(a[k] & (1 << i)); k++)
      ;
    if (k > n) continue;
    if (k ^ j) add(a, j, k, q), add(a, k, j, q), add(a, j, k, q);
    for (int k = 1; k <= n; k++)
      if ((a[k] & (1 << i)) && k != j) add(a, k, j, q);
    j++;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  work(a, qa);
  work(b, qb);
  for (int i = 1, k; i <= n && a[i]; i++) {
    for (k = 30; k + 1 && !((1 << k) & a[i]); k--)
      ;
    for (int j = 1; j <= 30 && j <= n; j++)
      if ((a[j] ^ b[j]) & (1 << k)) add(a, j, i, qa);
  }
  for (int i = 1; i <= n; i++)
    if (a[i] != b[i]) return puts("-1"), 0;
  printf("%d\n", qa.size() + qb.size());
  for (int i = 0; i < qa.size(); i++)
    printf("%d %d\n", qa[i].first, qa[i].second);
  for (int i = qb.size() - 1; i >= 0; i--)
    printf("%d %d\n", qb[i].first, qb[i].second);
}

### Prompt

Generate a Cpp solution to the following problem:

There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 10005;
int n, a[N], b[N];
vector<pair<int, int> > qa, qb;
void add(int *a, int i, int j, vector<pair<int, int> > &q) {
  a[i] ^= a[j];
  q.push_back(pair<int, int>(i, j));
}
void work(int *a, vector<pair<int, int> > &q) {
  for (int i = 30, j = 1, k; i >= 0; i--) {
    for (k = j; k <= n && !(a[k] & (1 << i)); k++)
      ;
    if (k > n) continue;
    if (k ^ j) add(a, j, k, q), add(a, k, j, q), add(a, j, k, q);
    for (int k = 1; k <= n; k++)
      if ((a[k] & (1 << i)) && k != j) add(a, k, j, q);
    j++;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  work(a, qa);
  work(b, qb);
  for (int i = 1, k; i <= n && a[i]; i++) {
    for (k = 30; k + 1 && !((1 << k) & a[i]); k--)
      ;
    for (int j = 1; j <= 30 && j <= n; j++)
      if ((a[j] ^ b[j]) & (1 << k)) add(a, j, i, qa);
  }
  for (int i = 1; i <= n; i++)
    if (a[i] != b[i]) return puts("-1"), 0;
  printf("%d\n", qa.size() + qb.size());
  for (int i = 0; i < qa.size(); i++)
    printf("%d %d\n", qa[i].first, qa[i].second);
  for (int i = qb.size() - 1; i >= 0; i--)
    printf("%d %d\n", qb[i].first, qb[i].second);
}

```