#include <bits/stdc++.h>
inline int read() {
  int ret;
  scanf("%d", &ret);
  return ret;
}
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
using namespace std;
int x[11111], y[11111];
int n;
vector<pair<int, int> > ans0, ans1;
int prework(int x[], vector<pair<int, int> > &ans) {
  int start = 1;
  for (int i = 30; i >= 0; i--) {
    for (int j = start; j <= n; j++) {
      if ((x[j] >> i) & 1) {
        swap(x[start], x[j]);
        if (start != j) {
          ans.push_back(make_pair(j, start));
          ans.push_back(make_pair(start, j));
          ans.push_back(make_pair(j, start));
        }
        for (int k = 1; k <= int(n); k++) {
          if (k != start && ((x[k] >> i) & 1)) {
            x[k] ^= x[start];
            ans.push_back(make_pair(k, start));
          }
        }
        start++;
        break;
      }
    }
  }
  return start - 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= int(n); i++) scanf("%d", &x[i]);
  for (int j = 1; j <= int(n); j++) scanf("%d", &y[j]);
  int l1 = prework(x, ans0);
  int l2 = prework(y, ans1);
  for (int i = 1; i <= int(l1); i++) {
    int c = 30;
    while (((x[i] >> c) & 1) == 0) c--;
    for (int j = 1; j <= int(l1); j++) {
      if (((y[j] >> c) & 1) != ((x[j] >> c) & 1)) {
        ans0.push_back(make_pair(j, i));
        x[j] ^= x[i];
      }
    }
  }
  bool YES = true;
  for (int i = 1; i <= int(l1); i++)
    if (x[i] != y[i]) YES = false;
  if (l1 < l2) YES = false;
  if (YES == false) {
    puts("-1");
    return 0;
  }
  printf("%d\n", ans0.size() + ans1.size());
  for (int i = 0; i < int(ans0.size()); i++)
    printf("%d %d\n", ans0[i].first, ans0[i].second);
  for (int i = 0; i < int(ans1.size()); i++)
    printf("%d %d\n", ans1[ans1.size() - i - 1].first,
           ans1[ans1.size() - i - 1].second);
  return 0;
}

### Prompt

Create a solution in cpp for 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>
inline int read() {
  int ret;
  scanf("%d", &ret);
  return ret;
}
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
using namespace std;
int x[11111], y[11111];
int n;
vector<pair<int, int> > ans0, ans1;
int prework(int x[], vector<pair<int, int> > &ans) {
  int start = 1;
  for (int i = 30; i >= 0; i--) {
    for (int j = start; j <= n; j++) {
      if ((x[j] >> i) & 1) {
        swap(x[start], x[j]);
        if (start != j) {
          ans.push_back(make_pair(j, start));
          ans.push_back(make_pair(start, j));
          ans.push_back(make_pair(j, start));
        }
        for (int k = 1; k <= int(n); k++) {
          if (k != start && ((x[k] >> i) & 1)) {
            x[k] ^= x[start];
            ans.push_back(make_pair(k, start));
          }
        }
        start++;
        break;
      }
    }
  }
  return start - 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= int(n); i++) scanf("%d", &x[i]);
  for (int j = 1; j <= int(n); j++) scanf("%d", &y[j]);
  int l1 = prework(x, ans0);
  int l2 = prework(y, ans1);
  for (int i = 1; i <= int(l1); i++) {
    int c = 30;
    while (((x[i] >> c) & 1) == 0) c--;
    for (int j = 1; j <= int(l1); j++) {
      if (((y[j] >> c) & 1) != ((x[j] >> c) & 1)) {
        ans0.push_back(make_pair(j, i));
        x[j] ^= x[i];
      }
    }
  }
  bool YES = true;
  for (int i = 1; i <= int(l1); i++)
    if (x[i] != y[i]) YES = false;
  if (l1 < l2) YES = false;
  if (YES == false) {
    puts("-1");
    return 0;
  }
  printf("%d\n", ans0.size() + ans1.size());
  for (int i = 0; i < int(ans0.size()); i++)
    printf("%d %d\n", ans0[i].first, ans0[i].second);
  for (int i = 0; i < int(ans1.size()); i++)
    printf("%d %d\n", ans1[ans1.size() - i - 1].first,
           ans1[ans1.size() - i - 1].second);
  return 0;
}

```