#include <bits/stdc++.h>
using namespace std;
vector<pair<long, long> > ops;
vector<pair<long, long> > restore;
long n, q;
vector<long> v1, v2;
void swap(long a, long b, vector<long>& v) {
  if (a == b) return;
  ops.push_back(make_pair(a, b));
  ops.push_back(make_pair(b, a));
  ops.push_back(make_pair(a, b));
  swap(v[a], v[b]);
}
void xorr(long a, long b, vector<long>& v) {
  ops.push_back(make_pair(a, b));
  v[a] ^= v[b];
}
void run_gauss(vector<long>& v) {
  long rows = 0;
  for (int bit = 30; bit >= 0; --bit) {
    for (int j = rows; j < v.size(); j++)
      if (v[j] & (1 << bit)) {
        swap(j, rows, v);
        for (int k = 0; k < v.size(); k++)
          if (k != rows && (v[k] & (1 << bit))) xorr(k, rows, v);
        ++rows;
        break;
      }
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> q;
    v1.push_back(q);
  }
  for (int i = 0; i < n; i++) {
    cin >> q;
    v2.push_back(q);
  }
  run_gauss(v2);
  restore = ops;
  ops.clear();
  run_gauss(v1);
  long er = 0;
  for (int i = 0; i < 40 && i < v1.size(); i++) {
    long old = 30;
    if (v1[i] == 0) continue;
    while ((v1[i] & (1 << old)) == 0) --old;
    for (int j = 0; j < 40 && j < v1.size() && j < v2.size(); j++)
      if ((v1[j] ^ v2[j]) & (1 << old)) xorr(j, i, v1);
  }
  for (int i = 0; i < v1.size(); i++)
    if (v1[i] != v2[i]) er = 1;
  reverse(restore.begin(), restore.end());
  for (int i = 0; i < restore.size(); i++) ops.push_back(restore[i]);
  if (er)
    cout << -1 << endl;
  else {
    cout << ops.size() << endl;
    for (int i = 0; i < ops.size(); i++)
      cout << ops[i].first + 1 << " " << ops[i].second + 1 << "\n";
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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<long, long> > ops;
vector<pair<long, long> > restore;
long n, q;
vector<long> v1, v2;
void swap(long a, long b, vector<long>& v) {
  if (a == b) return;
  ops.push_back(make_pair(a, b));
  ops.push_back(make_pair(b, a));
  ops.push_back(make_pair(a, b));
  swap(v[a], v[b]);
}
void xorr(long a, long b, vector<long>& v) {
  ops.push_back(make_pair(a, b));
  v[a] ^= v[b];
}
void run_gauss(vector<long>& v) {
  long rows = 0;
  for (int bit = 30; bit >= 0; --bit) {
    for (int j = rows; j < v.size(); j++)
      if (v[j] & (1 << bit)) {
        swap(j, rows, v);
        for (int k = 0; k < v.size(); k++)
          if (k != rows && (v[k] & (1 << bit))) xorr(k, rows, v);
        ++rows;
        break;
      }
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> q;
    v1.push_back(q);
  }
  for (int i = 0; i < n; i++) {
    cin >> q;
    v2.push_back(q);
  }
  run_gauss(v2);
  restore = ops;
  ops.clear();
  run_gauss(v1);
  long er = 0;
  for (int i = 0; i < 40 && i < v1.size(); i++) {
    long old = 30;
    if (v1[i] == 0) continue;
    while ((v1[i] & (1 << old)) == 0) --old;
    for (int j = 0; j < 40 && j < v1.size() && j < v2.size(); j++)
      if ((v1[j] ^ v2[j]) & (1 << old)) xorr(j, i, v1);
  }
  for (int i = 0; i < v1.size(); i++)
    if (v1[i] != v2[i]) er = 1;
  reverse(restore.begin(), restore.end());
  for (int i = 0; i < restore.size(); i++) ops.push_back(restore[i]);
  if (er)
    cout << -1 << endl;
  else {
    cout << ops.size() << endl;
    for (int i = 0; i < ops.size(); i++)
      cout << ops[i].first + 1 << " " << ops[i].second + 1 << "\n";
  }
  return 0;
}

```