#include <bits/stdc++.h>
using namespace std;
int N, A[1000005], B[1000005];
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(make_pair(i, j));
}
void Work(int *A, vector<pair<int, int> > &Q) {
  for (int i = 30, j = 1, k; i + 1; i--) {
    for (k = j; k <= N && !(A[k] & (1 << i)); k++)
      ;
    if (k > N) continue;
    if (k ^ j) Add(A, k, j, Q), Add(A, j, k, Q), Add(A, k, j, Q);
    for (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);
  int flag = 0;
  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]) flag = 1;
  if (flag)
    puts("-1");
  else {
    printf("%d\n", Qa.size() + Qb.size());
    for (int i = 0, s = Qa.size(); i < s; i++)
      printf("%d %d\n", Qa[i].first, Qa[i].second);
    for (int i = Qb.size() - 1; i + 1; i--)
      printf("%d %d\n", Qb[i].first, Qb[i].second);
  }
}

### Prompt

Please create a solution in CPP 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;
int N, A[1000005], B[1000005];
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(make_pair(i, j));
}
void Work(int *A, vector<pair<int, int> > &Q) {
  for (int i = 30, j = 1, k; i + 1; i--) {
    for (k = j; k <= N && !(A[k] & (1 << i)); k++)
      ;
    if (k > N) continue;
    if (k ^ j) Add(A, k, j, Q), Add(A, j, k, Q), Add(A, k, j, Q);
    for (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);
  int flag = 0;
  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]) flag = 1;
  if (flag)
    puts("-1");
  else {
    printf("%d\n", Qa.size() + Qb.size());
    for (int i = 0, s = Qa.size(); i < s; i++)
      printf("%d %d\n", Qa[i].first, Qa[i].second);
    for (int i = Qb.size() - 1; i + 1; i--)
      printf("%d %d\n", Qb[i].first, Qb[i].second);
  }
}

```