#include <bits/stdc++.h>
std::vector<std::pair<int, int>> vec1, vec2;
int n, x[10010], y[10010];
int a[33], posa[33];
int insert(int x, int d) {
  for (int i = 29; ~i; i--)
    if (x & (1ll << i)) {
      if (!a[i]) {
        a[i] = x, posa[i] = d;
        return i;
      } else
        x ^= a[i];
    }
  return -1;
}
int inserta(int x, int d) {
  for (int i = 29; ~i; i--)
    if (x & (1ll << i)) {
      if (!a[i]) {
        a[i] = x, posa[i] = d;
        return i;
      } else
        x ^= a[i], vec1.emplace_back(d, posa[i]);
    }
  return -1;
}
int b[33], posb[33];
int insertb(int x, int d) {
  for (int i = 29; ~i; i--)
    if (x & (1ll << i)) {
      if (!b[i]) {
        b[i] = x, posb[i] = d;
        return i;
      } else
        x ^= b[i], vec2.emplace_back(d, posb[i]);
    }
  return -1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", x + i);
    x[i] = inserta(x[i], i);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", y + i);
    y[i] = insertb(y[i], i);
  }
  for (int i = 0; i < 30; i++)
    if (~insert(b[i], 0)) return puts("-1"), 0;
  for (int i = 0; i < 30; i++)
    if (a[i] && b[i] && posa[i] != posb[i]) {
      vec1.emplace_back(posa[i], posb[i]);
      vec1.emplace_back(posb[i], posa[i]);
      vec1.emplace_back(posa[i], posb[i]);
      std::swap(x[posb[i]], x[posa[i]]);
      if (~x[posa[i]])
        std::swap(posa[x[posa[i]]], posa[i]);
      else
        posa[i] = posb[i];
    }
  for (int i = 29; ~i; i--)
    if (a[i]) inserta(a[i] ^ b[i], posa[i]);
  printf("%d\n", int(vec1.size() + vec2.size()));
  for (auto u : vec1) printf("%d %d\n", u.first, u.second);
  std::reverse(vec2.begin(), vec2.end());
  for (auto u : vec2) printf("%d %d\n", u.first, u.second);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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>
std::vector<std::pair<int, int>> vec1, vec2;
int n, x[10010], y[10010];
int a[33], posa[33];
int insert(int x, int d) {
  for (int i = 29; ~i; i--)
    if (x & (1ll << i)) {
      if (!a[i]) {
        a[i] = x, posa[i] = d;
        return i;
      } else
        x ^= a[i];
    }
  return -1;
}
int inserta(int x, int d) {
  for (int i = 29; ~i; i--)
    if (x & (1ll << i)) {
      if (!a[i]) {
        a[i] = x, posa[i] = d;
        return i;
      } else
        x ^= a[i], vec1.emplace_back(d, posa[i]);
    }
  return -1;
}
int b[33], posb[33];
int insertb(int x, int d) {
  for (int i = 29; ~i; i--)
    if (x & (1ll << i)) {
      if (!b[i]) {
        b[i] = x, posb[i] = d;
        return i;
      } else
        x ^= b[i], vec2.emplace_back(d, posb[i]);
    }
  return -1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", x + i);
    x[i] = inserta(x[i], i);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", y + i);
    y[i] = insertb(y[i], i);
  }
  for (int i = 0; i < 30; i++)
    if (~insert(b[i], 0)) return puts("-1"), 0;
  for (int i = 0; i < 30; i++)
    if (a[i] && b[i] && posa[i] != posb[i]) {
      vec1.emplace_back(posa[i], posb[i]);
      vec1.emplace_back(posb[i], posa[i]);
      vec1.emplace_back(posa[i], posb[i]);
      std::swap(x[posb[i]], x[posa[i]]);
      if (~x[posa[i]])
        std::swap(posa[x[posa[i]]], posa[i]);
      else
        posa[i] = posb[i];
    }
  for (int i = 29; ~i; i--)
    if (a[i]) inserta(a[i] ^ b[i], posa[i]);
  printf("%d\n", int(vec1.size() + vec2.size()));
  for (auto u : vec1) printf("%d %d\n", u.first, u.second);
  std::reverse(vec2.begin(), vec2.end());
  for (auto u : vec2) printf("%d %d\n", u.first, u.second);
  return 0;
}

```