#include <bits/stdc++.h>
const int MAXN = 10101;
const int MAXB = 31;
int arr[2][MAXN];
int n;
inline int bit(int i, int j, int k) { return arr[i][j] >> k & 1; }
struct Pair {
  int a, b;
  Pair(int aa = 0, int ba = 0) : a(aa + 1), b(ba + 1) {}
  void print() const { printf("%d %d\n", a, b); }
};
std::vector<Pair> ans[2];
inline void add(int i, int j, int k) {
  ans[i].push_back(Pair(j, k));
  arr[i][j] ^= arr[i][k];
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < 2; ++i) {
    for (int j = 0; j < n; ++j) {
      scanf("%d", &arr[i][j]);
    }
  }
  int at[2][MAXB];
  int lat[2] = {0};
  for (int i = 0; i < 2; ++i) {
    int &L = lat[i];
    for (int j = 0; j < MAXB; ++j) {
      for (int k = L + 1; k < n; ++k) {
        if (bit(i, k, j)) {
          add(i, L, k);
          add(i, k, L);
          add(i, L, k);
          break;
        }
      }
      if (not bit(i, L, j)) {
        at[i][j] = -1;
        continue;
      }
      for (int k = 0; k < n; ++k)
        if (bit(i, k, j) and k != L) add(i, k, L);
      at[i][j] = L++;
    }
  }
  for (int i = 0; i < lat[0]; ++i)
    for (int j = 0; j < MAXB; ++j)
      if (bit(0, i, j) != bit(1, i, j) and at[0][j] != -1) add(0, i, at[0][j]);
  for (int i = 0; i < n; ++i) {
    if (arr[0][i] != arr[1][i]) {
      puts("-1");
      return 0;
    }
  }
  printf("%d\n", ans[0].size() + ans[1].size());
  for (int i = 0; i < (int)ans[0].size(); ++i) ans[0][i].print();
  for (int i = (int)ans[1].size() - 1; i >= 0; --i) ans[1][i].print();
  return 0;
}

### Prompt

Please formulate 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>
const int MAXN = 10101;
const int MAXB = 31;
int arr[2][MAXN];
int n;
inline int bit(int i, int j, int k) { return arr[i][j] >> k & 1; }
struct Pair {
  int a, b;
  Pair(int aa = 0, int ba = 0) : a(aa + 1), b(ba + 1) {}
  void print() const { printf("%d %d\n", a, b); }
};
std::vector<Pair> ans[2];
inline void add(int i, int j, int k) {
  ans[i].push_back(Pair(j, k));
  arr[i][j] ^= arr[i][k];
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < 2; ++i) {
    for (int j = 0; j < n; ++j) {
      scanf("%d", &arr[i][j]);
    }
  }
  int at[2][MAXB];
  int lat[2] = {0};
  for (int i = 0; i < 2; ++i) {
    int &L = lat[i];
    for (int j = 0; j < MAXB; ++j) {
      for (int k = L + 1; k < n; ++k) {
        if (bit(i, k, j)) {
          add(i, L, k);
          add(i, k, L);
          add(i, L, k);
          break;
        }
      }
      if (not bit(i, L, j)) {
        at[i][j] = -1;
        continue;
      }
      for (int k = 0; k < n; ++k)
        if (bit(i, k, j) and k != L) add(i, k, L);
      at[i][j] = L++;
    }
  }
  for (int i = 0; i < lat[0]; ++i)
    for (int j = 0; j < MAXB; ++j)
      if (bit(0, i, j) != bit(1, i, j) and at[0][j] != -1) add(0, i, at[0][j]);
  for (int i = 0; i < n; ++i) {
    if (arr[0][i] != arr[1][i]) {
      puts("-1");
      return 0;
    }
  }
  printf("%d\n", ans[0].size() + ans[1].size());
  for (int i = 0; i < (int)ans[0].size(); ++i) ans[0][i].print();
  for (int i = (int)ans[1].size() - 1; i >= 0; --i) ans[1][i].print();
  return 0;
}

```