#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n;
  scanf("%d", &n);
  vector<int> a(n), b(n), pa, pb;
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  pa = a;
  for (int i = 0; i < n; ++i) scanf("%d", &b[i]);
  pb = b;
  vector<pair<int, int> > aa, bb;
  int i, j, k, q = 0;
  int msk[60];
  for (j = 32; j >= 0; j--) {
    msk[j] = -1;
    for (int i = q; i < n; i++) {
      if (a[i] & (1LL << j)) {
        msk[j] = q;
        if (i != q) {
          swap(a[i], a[q]);
          aa.push_back(make_pair(i, q));
          aa.push_back(make_pair(q, i));
          aa.push_back(make_pair(i, q));
        }
        q++;
        for (k = q; k < n; k++)
          if (a[k] & (1LL << j)) {
            a[k] ^= a[q - 1];
            aa.push_back(make_pair(k, q - 1));
          }
      }
    }
  }
  i = j = k = q = 0;
  for (j = 32; j >= 0; j--) {
    for (i = q; i < n; i++)
      if (b[i] & (1LL << j)) {
        if (i != q) {
          swap(b[i], b[q]);
          bb.push_back(make_pair(i, q));
          bb.push_back(make_pair(q, i));
          bb.push_back(make_pair(i, q));
        }
        q++;
        for (k = q; k < n; k++)
          if (b[k] & (1LL << j)) {
            b[k] ^= b[q - 1];
            bb.push_back(make_pair(k, q - 1));
          }
      }
  }
  vector<int> mt(n);
  i = 0, j = 0;
  for (int j = 0; j < n; ++j) {
    bool is_used = false;
    int t = b[j];
    int cur = 0;
    for (int i = 31; i >= 0; --i)
      if (((t >> i) & 1) != ((cur >> i) & 1)) {
        if (msk[i] == -1) {
          cout << "-1\n";
          return;
        }
        if (!is_used) {
          if (j != msk[i]) {
            aa.push_back(make_pair(j, msk[i]));
            aa.push_back(make_pair(msk[i], j));
            aa.push_back(make_pair(j, msk[i]));
          }
          is_used = true;
        } else
          aa.push_back(make_pair(j, msk[i]));
        cur ^= a[msk[i]];
      }
    if (b[j] == 0) aa.push_back(make_pair(j, j));
  }
  for (int i = bb.size() - 1; i >= 0; --i) aa.push_back(bb[i]);
  printf("%d\n", aa.size());
  for (int i = 0; i < aa.size(); ++i) {
    printf("%d %d\n", aa[i].first + 1, aa[i].second + 1);
    pa[aa[i].first] ^= pa[aa[i].second];
  }
}
int main() {
  std::ios_base::sync_with_stdio(false);
  solve();
  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;
void solve() {
  int n;
  scanf("%d", &n);
  vector<int> a(n), b(n), pa, pb;
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  pa = a;
  for (int i = 0; i < n; ++i) scanf("%d", &b[i]);
  pb = b;
  vector<pair<int, int> > aa, bb;
  int i, j, k, q = 0;
  int msk[60];
  for (j = 32; j >= 0; j--) {
    msk[j] = -1;
    for (int i = q; i < n; i++) {
      if (a[i] & (1LL << j)) {
        msk[j] = q;
        if (i != q) {
          swap(a[i], a[q]);
          aa.push_back(make_pair(i, q));
          aa.push_back(make_pair(q, i));
          aa.push_back(make_pair(i, q));
        }
        q++;
        for (k = q; k < n; k++)
          if (a[k] & (1LL << j)) {
            a[k] ^= a[q - 1];
            aa.push_back(make_pair(k, q - 1));
          }
      }
    }
  }
  i = j = k = q = 0;
  for (j = 32; j >= 0; j--) {
    for (i = q; i < n; i++)
      if (b[i] & (1LL << j)) {
        if (i != q) {
          swap(b[i], b[q]);
          bb.push_back(make_pair(i, q));
          bb.push_back(make_pair(q, i));
          bb.push_back(make_pair(i, q));
        }
        q++;
        for (k = q; k < n; k++)
          if (b[k] & (1LL << j)) {
            b[k] ^= b[q - 1];
            bb.push_back(make_pair(k, q - 1));
          }
      }
  }
  vector<int> mt(n);
  i = 0, j = 0;
  for (int j = 0; j < n; ++j) {
    bool is_used = false;
    int t = b[j];
    int cur = 0;
    for (int i = 31; i >= 0; --i)
      if (((t >> i) & 1) != ((cur >> i) & 1)) {
        if (msk[i] == -1) {
          cout << "-1\n";
          return;
        }
        if (!is_used) {
          if (j != msk[i]) {
            aa.push_back(make_pair(j, msk[i]));
            aa.push_back(make_pair(msk[i], j));
            aa.push_back(make_pair(j, msk[i]));
          }
          is_used = true;
        } else
          aa.push_back(make_pair(j, msk[i]));
        cur ^= a[msk[i]];
      }
    if (b[j] == 0) aa.push_back(make_pair(j, j));
  }
  for (int i = bb.size() - 1; i >= 0; --i) aa.push_back(bb[i]);
  printf("%d\n", aa.size());
  for (int i = 0; i < aa.size(); ++i) {
    printf("%d %d\n", aa[i].first + 1, aa[i].second + 1);
    pa[aa[i].first] ^= pa[aa[i].second];
  }
}
int main() {
  std::ios_base::sync_with_stdio(false);
  solve();
  return 0;
}

```