#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > changes;
vector<int> norm(vector<int> x) {
  int i = -1;
  for (int bit = 30; bit >= 0; bit--) {
    for (int pos = i + 1; pos < x.size(); pos++)
      if (x[pos] & (1 << bit)) {
        if (pos != ++i) {
          changes.push_back(pair<int, int>(pos, i));
          x[pos] ^= x[i];
          changes.push_back(pair<int, int>(i, pos));
          x[i] ^= x[pos];
        }
        for (int pos = 0; pos < x.size(); pos++)
          if (pos != i) {
            if (x[pos] & (1 << bit)) {
              changes.push_back(pair<int, int>(pos, i));
              x[pos] ^= x[i];
            }
          }
        break;
      }
  }
  return x;
}
int used[33];
int main() {
  int n;
  cin >> n;
  vector<int> x(n), y(n);
  for (int i = 0; i < n; i++) scanf("%d", &x[i]);
  for (int i = 0; i < n; i++) scanf("%d", &y[i]);
  vector<pair<int, int> > a, b;
  vector<int> bx = norm(x);
  a = changes;
  changes.clear();
  vector<int> by = norm(y);
  b = changes;
  reverse(b.begin(), b.end());
  for (int i = 0; i < n; i++)
    if (by[i]) used[__builtin_clz(by[i])] = 1;
  for (int i = 0; i < n; i++)
    if (bx[i] && used[__builtin_clz(bx[i])]) {
      for (int j = i - 1; j >= 0; j--) {
        if (bx[j] && used[__builtin_clz(bx[j])]) break;
        a.push_back(pair<int, int>(j, j + 1));
        bx[j + 1] ^= bx[j];
        a.push_back(pair<int, int>(j + 1, j));
        bx[j] ^= bx[j + 1];
        a.push_back(pair<int, int>(j, j + 1));
        bx[j + 1] ^= bx[j];
      }
    }
  for (int i = 0; i < n; i++) {
    if (by[i] != bx[i]) {
      if (!by[i]) {
        bx[i] = 0;
        a.push_back(pair<int, int>(i, i));
      } else {
        for (int j = i + 1; j < n; j++)
          if (bx[j] && (by[i] & (1 << (31 - __builtin_clz(bx[j]))))) {
            bx[i] ^= bx[j];
            a.push_back(pair<int, int>(i, j));
          }
        if (by[i] != bx[i]) {
          cout << -1 << endl;
          return 0;
        }
      }
    }
  }
  vector<int> aux = x;
  cout << a.size() + b.size() << endl;
  for (int i = 0; i < a.size(); i++) {
    cout << a[i].first + 1 << " " << a[i].second + 1 << endl;
    aux[a[i].first] ^= aux[a[i].second];
  }
  for (int i = 0; i < b.size(); i++) {
    cout << b[i].first + 1 << " " << b[i].second + 1 << endl;
    aux[b[i].first] ^= aux[b[i].second];
  }
  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>
using namespace std;
vector<pair<int, int> > changes;
vector<int> norm(vector<int> x) {
  int i = -1;
  for (int bit = 30; bit >= 0; bit--) {
    for (int pos = i + 1; pos < x.size(); pos++)
      if (x[pos] & (1 << bit)) {
        if (pos != ++i) {
          changes.push_back(pair<int, int>(pos, i));
          x[pos] ^= x[i];
          changes.push_back(pair<int, int>(i, pos));
          x[i] ^= x[pos];
        }
        for (int pos = 0; pos < x.size(); pos++)
          if (pos != i) {
            if (x[pos] & (1 << bit)) {
              changes.push_back(pair<int, int>(pos, i));
              x[pos] ^= x[i];
            }
          }
        break;
      }
  }
  return x;
}
int used[33];
int main() {
  int n;
  cin >> n;
  vector<int> x(n), y(n);
  for (int i = 0; i < n; i++) scanf("%d", &x[i]);
  for (int i = 0; i < n; i++) scanf("%d", &y[i]);
  vector<pair<int, int> > a, b;
  vector<int> bx = norm(x);
  a = changes;
  changes.clear();
  vector<int> by = norm(y);
  b = changes;
  reverse(b.begin(), b.end());
  for (int i = 0; i < n; i++)
    if (by[i]) used[__builtin_clz(by[i])] = 1;
  for (int i = 0; i < n; i++)
    if (bx[i] && used[__builtin_clz(bx[i])]) {
      for (int j = i - 1; j >= 0; j--) {
        if (bx[j] && used[__builtin_clz(bx[j])]) break;
        a.push_back(pair<int, int>(j, j + 1));
        bx[j + 1] ^= bx[j];
        a.push_back(pair<int, int>(j + 1, j));
        bx[j] ^= bx[j + 1];
        a.push_back(pair<int, int>(j, j + 1));
        bx[j + 1] ^= bx[j];
      }
    }
  for (int i = 0; i < n; i++) {
    if (by[i] != bx[i]) {
      if (!by[i]) {
        bx[i] = 0;
        a.push_back(pair<int, int>(i, i));
      } else {
        for (int j = i + 1; j < n; j++)
          if (bx[j] && (by[i] & (1 << (31 - __builtin_clz(bx[j]))))) {
            bx[i] ^= bx[j];
            a.push_back(pair<int, int>(i, j));
          }
        if (by[i] != bx[i]) {
          cout << -1 << endl;
          return 0;
        }
      }
    }
  }
  vector<int> aux = x;
  cout << a.size() + b.size() << endl;
  for (int i = 0; i < a.size(); i++) {
    cout << a[i].first + 1 << " " << a[i].second + 1 << endl;
    aux[a[i].first] ^= aux[a[i].second];
  }
  for (int i = 0; i < b.size(); i++) {
    cout << b[i].first + 1 << " " << b[i].second + 1 << endl;
    aux[b[i].first] ^= aux[b[i].second];
  }
  return 0;
}

```