#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > ops, res;
vector<long> a, b;
int n;
long long x;
void swap(int i, int j, vector<long> &v) {
  if (i == j) return;
  ops.push_back(make_pair(i, j));
  ops.push_back(make_pair(j, i));
  ops.push_back(make_pair(i, j));
  swap(v[i], v[j]);
}
void xorr(int i, int j, vector<long> &v) {
  v[i] ^= v[j];
  ops.push_back(make_pair(i, j));
}
void gauss(vector<long> &v) {
  int row = 0;
  for (register int j = 30; j >= 0; --j) {
    for (register int i = row; i <= n - 1; ++i) {
      if (v[i] & (1 << j)) {
        swap(i, row, v);
        for (register int k = 0; k <= n - 1; ++k)
          if (k != row)
            if (v[k] & (1 << j)) xorr(k, row, v);
        row++;
        break;
      }
    }
  }
}
int main() {
  cin >> n;
  for (register int i = 0; i <= n - 1; ++i) {
    cin >> x;
    a.push_back(x);
  }
  for (register int i = 0; i <= n - 1; ++i) {
    cin >> x;
    b.push_back(x);
  }
  gauss(b);
  res = ops;
  ops.clear();
  gauss(a);
  for (int i = 0; i < n && i < 35; ++i) {
    int x = 30;
    if (!a[i]) continue;
    while (!(a[i] & (1 << x))) --x;
    for (int j = 0; j < n && j < 35; ++j)
      if ((a[j] ^ b[j]) & (1 << x)) xorr(j, i, a);
  }
  for (register int i = 0; i <= n - 1; ++i)
    if (a[i] != b[i]) {
      cout << "-1";
      return 0;
    }
  reverse(res.begin(), res.end());
  for (vector<pair<int, int> >::iterator it = res.begin(); it != res.end();
       it++)
    ops.push_back(*it);
  cout << ops.size() << "\n";
  for (vector<pair<int, int> >::iterator it = ops.begin(); it != ops.end();
       it++)
    cout << it->first + 1 << " " << it->second + 1 << "\n";
  return 0;
}

### Prompt

Your challenge is to write 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> > ops, res;
vector<long> a, b;
int n;
long long x;
void swap(int i, int j, vector<long> &v) {
  if (i == j) return;
  ops.push_back(make_pair(i, j));
  ops.push_back(make_pair(j, i));
  ops.push_back(make_pair(i, j));
  swap(v[i], v[j]);
}
void xorr(int i, int j, vector<long> &v) {
  v[i] ^= v[j];
  ops.push_back(make_pair(i, j));
}
void gauss(vector<long> &v) {
  int row = 0;
  for (register int j = 30; j >= 0; --j) {
    for (register int i = row; i <= n - 1; ++i) {
      if (v[i] & (1 << j)) {
        swap(i, row, v);
        for (register int k = 0; k <= n - 1; ++k)
          if (k != row)
            if (v[k] & (1 << j)) xorr(k, row, v);
        row++;
        break;
      }
    }
  }
}
int main() {
  cin >> n;
  for (register int i = 0; i <= n - 1; ++i) {
    cin >> x;
    a.push_back(x);
  }
  for (register int i = 0; i <= n - 1; ++i) {
    cin >> x;
    b.push_back(x);
  }
  gauss(b);
  res = ops;
  ops.clear();
  gauss(a);
  for (int i = 0; i < n && i < 35; ++i) {
    int x = 30;
    if (!a[i]) continue;
    while (!(a[i] & (1 << x))) --x;
    for (int j = 0; j < n && j < 35; ++j)
      if ((a[j] ^ b[j]) & (1 << x)) xorr(j, i, a);
  }
  for (register int i = 0; i <= n - 1; ++i)
    if (a[i] != b[i]) {
      cout << "-1";
      return 0;
    }
  reverse(res.begin(), res.end());
  for (vector<pair<int, int> >::iterator it = res.begin(); it != res.end();
       it++)
    ops.push_back(*it);
  cout << ops.size() << "\n";
  for (vector<pair<int, int> >::iterator it = ops.begin(); it != ops.end();
       it++)
    cout << it->first + 1 << " " << it->second + 1 << "\n";
  return 0;
}

```