#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > AA, BB;
int a[20000], b[20000], n, flag;
void change(int *a, int i, int j, vector<pair<int, int> > &Q) {
  Q.push_back(pair<int, int>(i, j));
  a[i] ^= a[j];
}
void work(int *a, vector<pair<int, int> > &Q) {
  int r = 0;
  for (int i = 30; i >= 0; --i)
    for (int j = r; j < n; ++j)
      if ((a[j] & (1 << i))) {
        if (j != r) {
          change(a, r, j, Q);
          change(a, j, r, Q);
          change(a, r, j, Q);
        }
        for (int k = 0; k < n; ++k)
          if (k != r && (a[k] & (1 << i))) change(a, k, r, Q);
        r++;
        break;
      }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  for (int i = 0; i < n; ++i) scanf("%d", &b[i]);
  work(a, AA);
  work(b, BB);
  for (int i = 0; i < n; ++i) {
    if (a[i] == 0) break;
    int dq = 1 << 30;
    while (!(dq & a[i])) dq >>= 1;
    for (int j = 0; j <= 30 && j < n; ++j)
      if ((a[j] ^ b[j]) & dq) change(a, j, i, AA);
  }
  flag = 1;
  for (int i = 0; i < n; ++i)
    if (a[i] != b[i]) flag = 0;
  if (!flag) {
    printf("-1\n");
    return 0;
  }
  printf("%d\n", AA.size() + BB.size());
  for (int i = 0; i < AA.size(); ++i)
    printf("%d %d\n", AA[i].first + 1, AA[i].second + 1);
  reverse(BB.begin(), BB.end());
  for (int i = 0; i < BB.size(); ++i)
    printf("%d %d\n", BB[i].first + 1, BB[i].second + 1);
  return 0;
}

### 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;
vector<pair<int, int> > AA, BB;
int a[20000], b[20000], n, flag;
void change(int *a, int i, int j, vector<pair<int, int> > &Q) {
  Q.push_back(pair<int, int>(i, j));
  a[i] ^= a[j];
}
void work(int *a, vector<pair<int, int> > &Q) {
  int r = 0;
  for (int i = 30; i >= 0; --i)
    for (int j = r; j < n; ++j)
      if ((a[j] & (1 << i))) {
        if (j != r) {
          change(a, r, j, Q);
          change(a, j, r, Q);
          change(a, r, j, Q);
        }
        for (int k = 0; k < n; ++k)
          if (k != r && (a[k] & (1 << i))) change(a, k, r, Q);
        r++;
        break;
      }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  for (int i = 0; i < n; ++i) scanf("%d", &b[i]);
  work(a, AA);
  work(b, BB);
  for (int i = 0; i < n; ++i) {
    if (a[i] == 0) break;
    int dq = 1 << 30;
    while (!(dq & a[i])) dq >>= 1;
    for (int j = 0; j <= 30 && j < n; ++j)
      if ((a[j] ^ b[j]) & dq) change(a, j, i, AA);
  }
  flag = 1;
  for (int i = 0; i < n; ++i)
    if (a[i] != b[i]) flag = 0;
  if (!flag) {
    printf("-1\n");
    return 0;
  }
  printf("%d\n", AA.size() + BB.size());
  for (int i = 0; i < AA.size(); ++i)
    printf("%d %d\n", AA[i].first + 1, AA[i].second + 1);
  reverse(BB.begin(), BB.end());
  for (int i = 0; i < BB.size(); ++i)
    printf("%d %d\n", BB[i].first + 1, BB[i].second + 1);
  return 0;
}

```