#include <bits/stdc++.h>
using namespace std;
const int maxn = 10000 + 10;
int n;
int x[maxn], y[maxn];
vector<pair<int, int> > ans, ans2, ansMiddle;
int main(int argc, char *argv[]) {
  cin >> n;
  for (int i = (1); i <= (n); i++) scanf("%d", x + i);
  int p = 1;
  for (int i = (29); i >= (0); i--) {
    for (int j = (p); j <= (n); j++)
      if (x[j] & (1 << i)) {
        if (j == p) break;
        swap(x[p], x[j]);
        ans.push_back(make_pair(j, p));
        ans.push_back(make_pair(p, j));
        ans.push_back(make_pair(j, p));
        break;
      }
    if (x[p] & (1 << i)) {
      for (int j = (1); j <= (n); j++)
        if (p != j && (x[j] & (1 << i))) {
          x[j] ^= x[p];
          ans.push_back(make_pair(j, p));
        }
      p++;
      if (p == n + 1) break;
    }
  }
  int sumBase = p - 1;
  for (int k = (1); k <= (n); k++) scanf("%d", y + k);
  p = 1;
  for (int i = (29); i >= (0); i--) {
    for (int j = (p); j <= (n); j++)
      if (y[j] & (1 << i)) {
        if (j == p) break;
        swap(y[p], y[j]);
        ans2.push_back(make_pair(j, p));
        ans2.push_back(make_pair(p, j));
        ans2.push_back(make_pair(j, p));
        break;
      }
    if (y[p] & (1 << i)) {
      for (int j = (1); j <= (n); j++)
        if (p != j && (y[j] & (1 << i))) {
          y[j] ^= y[p];
          ans2.push_back(make_pair(j, p));
        }
      p++;
      if (p == n + 1) break;
    }
  }
  reverse(ans2.begin(), ans2.end());
  int wei[33];
  for (int i = (1); i <= (sumBase); i++) {
    int j = 29;
    while (!((1 << j) & x[i])) j--;
    wei[i] = j;
  }
  for (int i = (1); i <= (n); i++) {
    if (y[i] == 0) {
      ansMiddle.push_back(make_pair(i, i));
      continue;
    }
    vector<int> a;
    while (y[i] != 0) {
      int j = 29;
      while (!((1 << j) & y[i])) j--;
      bool find = false;
      for (int k = (1); k <= (sumBase); k++)
        if (wei[k] == j) {
          find = true;
          y[i] ^= x[k];
          a.push_back(k);
          break;
        }
      if (!find) {
        printf("-1\n");
        exit(0);
      }
    }
    if (a[0] != i) {
      ansMiddle.push_back(make_pair(i, i));
      ansMiddle.push_back(make_pair(i, a[0]));
    }
    for (int k = 1; k < ((int)(a).size()); k++)
      ansMiddle.push_back(make_pair(i, a[k]));
  }
  cout << ((int)(ans).size()) + ((int)(ans2).size()) + ((int)(ansMiddle).size())
       << "\n";
  for (typeof((ans).begin()) it = (ans).begin(); it != (ans).end(); it++)
    printf("%d %d\n", it->first, it->second);
  for (typeof((ansMiddle).begin()) it = (ansMiddle).begin();
       it != (ansMiddle).end(); it++)
    printf("%d %d\n", it->first, it->second);
  for (typeof((ans2).begin()) it = (ans2).begin(); it != (ans2).end(); it++)
    printf("%d %d\n", it->first, it->second);
  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;
const int maxn = 10000 + 10;
int n;
int x[maxn], y[maxn];
vector<pair<int, int> > ans, ans2, ansMiddle;
int main(int argc, char *argv[]) {
  cin >> n;
  for (int i = (1); i <= (n); i++) scanf("%d", x + i);
  int p = 1;
  for (int i = (29); i >= (0); i--) {
    for (int j = (p); j <= (n); j++)
      if (x[j] & (1 << i)) {
        if (j == p) break;
        swap(x[p], x[j]);
        ans.push_back(make_pair(j, p));
        ans.push_back(make_pair(p, j));
        ans.push_back(make_pair(j, p));
        break;
      }
    if (x[p] & (1 << i)) {
      for (int j = (1); j <= (n); j++)
        if (p != j && (x[j] & (1 << i))) {
          x[j] ^= x[p];
          ans.push_back(make_pair(j, p));
        }
      p++;
      if (p == n + 1) break;
    }
  }
  int sumBase = p - 1;
  for (int k = (1); k <= (n); k++) scanf("%d", y + k);
  p = 1;
  for (int i = (29); i >= (0); i--) {
    for (int j = (p); j <= (n); j++)
      if (y[j] & (1 << i)) {
        if (j == p) break;
        swap(y[p], y[j]);
        ans2.push_back(make_pair(j, p));
        ans2.push_back(make_pair(p, j));
        ans2.push_back(make_pair(j, p));
        break;
      }
    if (y[p] & (1 << i)) {
      for (int j = (1); j <= (n); j++)
        if (p != j && (y[j] & (1 << i))) {
          y[j] ^= y[p];
          ans2.push_back(make_pair(j, p));
        }
      p++;
      if (p == n + 1) break;
    }
  }
  reverse(ans2.begin(), ans2.end());
  int wei[33];
  for (int i = (1); i <= (sumBase); i++) {
    int j = 29;
    while (!((1 << j) & x[i])) j--;
    wei[i] = j;
  }
  for (int i = (1); i <= (n); i++) {
    if (y[i] == 0) {
      ansMiddle.push_back(make_pair(i, i));
      continue;
    }
    vector<int> a;
    while (y[i] != 0) {
      int j = 29;
      while (!((1 << j) & y[i])) j--;
      bool find = false;
      for (int k = (1); k <= (sumBase); k++)
        if (wei[k] == j) {
          find = true;
          y[i] ^= x[k];
          a.push_back(k);
          break;
        }
      if (!find) {
        printf("-1\n");
        exit(0);
      }
    }
    if (a[0] != i) {
      ansMiddle.push_back(make_pair(i, i));
      ansMiddle.push_back(make_pair(i, a[0]));
    }
    for (int k = 1; k < ((int)(a).size()); k++)
      ansMiddle.push_back(make_pair(i, a[k]));
  }
  cout << ((int)(ans).size()) + ((int)(ans2).size()) + ((int)(ansMiddle).size())
       << "\n";
  for (typeof((ans).begin()) it = (ans).begin(); it != (ans).end(); it++)
    printf("%d %d\n", it->first, it->second);
  for (typeof((ansMiddle).begin()) it = (ansMiddle).begin();
       it != (ansMiddle).end(); it++)
    printf("%d %d\n", it->first, it->second);
  for (typeof((ans2).begin()) it = (ans2).begin(); it != (ans2).end(); it++)
    printf("%d %d\n", it->first, it->second);
  return 0;
}

```