#include <bits/stdc++.h>
using namespace std;
int n;
vector<pair<int, int> > gauss(vector<int> &v) {
  vector<pair<int, int> > log;
  for (int i = 30, pos = 0; i >= 0; --i) {
    int idx = -1;
    for (int j = pos; j < n; ++j) {
      if (v[j] & (1 << i)) {
        idx = j;
        break;
      }
    }
    if (idx == -1) continue;
    if (idx != pos) {
      v[pos] ^= v[idx];
      log.push_back(pair<int, int>(pos, idx));
    }
    for (int j = 0; j < n; ++j) {
      if (j == pos) continue;
      if (v[j] & (1 << i)) {
        v[j] ^= v[pos];
        log.push_back(pair<int, int>(j, pos));
      }
    }
    ++pos;
  }
  return log;
}
int highest_bit(int p) {
  p |= (p >> 1);
  p |= (p >> 2);
  p |= (p >> 4);
  p |= (p >> 8);
  p |= (p >> 16);
  return (p + 1) >> 1;
}
int main(void) {
  cin >> n;
  vector<int> x(n), y(n);
  for (int i = 0; i < n; ++i) {
    cin >> x[i];
  }
  for (int i = 0; i < n; ++i) {
    cin >> y[i];
  }
  vector<int> orig_x = x, orig_y = y;
  vector<pair<int, int> > x_log = gauss(x);
  vector<pair<int, int> > y_log = gauss(y);
  vector<pair<int, int> > merge_log;
  for (int i = 0, j = 0; i < n; ++i) {
    if (y[i] == 0) {
      for (; j < n; ++j) {
        if (x[j] != 0) {
          merge_log.push_back(pair<int, int>(j, j));
          x[j] ^= x[j];
        }
      }
      break;
    }
    while (j < n && highest_bit(x[j]) > highest_bit(y[i])) {
      merge_log.push_back(pair<int, int>(j, j));
      x[j] ^= x[j];
      ++j;
    }
    if (j == n || highest_bit(x[j]) < highest_bit(y[i])) {
      cout << -1 << endl;
      return 0;
    }
    for (int k = j + 1; k < n && x[j] != y[i]; ++k) {
      int b1 = highest_bit(x[k]);
      int b2 = highest_bit(x[j] ^ y[i]);
      if (b2 > b1) {
        break;
      }
      if (b1 == b2) {
        x[j] ^= x[k];
        merge_log.push_back(pair<int, int>(j, k));
      }
    }
    if (x[j] != y[i]) {
      cout << -1 << endl;
      return 0;
    }
    if (j != i) {
      merge_log.push_back(pair<int, int>(j, i));
      x[j] ^= x[i];
      merge_log.push_back(pair<int, int>(i, j));
      x[i] ^= x[j];
      merge_log.push_back(pair<int, int>(j, j));
      x[j] ^= x[j];
    }
    ++j;
  }
  vector<pair<int, int> > result;
  result.insert(result.end(), x_log.begin(), x_log.end());
  result.insert(result.end(), merge_log.begin(), merge_log.end());
  result.insert(result.end(), y_log.rbegin(), y_log.rend());
  x = orig_x;
  y = orig_y;
  cout << result.size() << endl;
  for (int i = 0; i < (int)result.size(); ++i) {
    x[result[i].first] ^= x[result[i].second];
    cout << result[i].first + 1 << " " << result[i].second + 1 << endl;
  }
  if (x != y) {
    return 1;
  }
  return 0;
}

### Prompt

Your task is to create 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;
int n;
vector<pair<int, int> > gauss(vector<int> &v) {
  vector<pair<int, int> > log;
  for (int i = 30, pos = 0; i >= 0; --i) {
    int idx = -1;
    for (int j = pos; j < n; ++j) {
      if (v[j] & (1 << i)) {
        idx = j;
        break;
      }
    }
    if (idx == -1) continue;
    if (idx != pos) {
      v[pos] ^= v[idx];
      log.push_back(pair<int, int>(pos, idx));
    }
    for (int j = 0; j < n; ++j) {
      if (j == pos) continue;
      if (v[j] & (1 << i)) {
        v[j] ^= v[pos];
        log.push_back(pair<int, int>(j, pos));
      }
    }
    ++pos;
  }
  return log;
}
int highest_bit(int p) {
  p |= (p >> 1);
  p |= (p >> 2);
  p |= (p >> 4);
  p |= (p >> 8);
  p |= (p >> 16);
  return (p + 1) >> 1;
}
int main(void) {
  cin >> n;
  vector<int> x(n), y(n);
  for (int i = 0; i < n; ++i) {
    cin >> x[i];
  }
  for (int i = 0; i < n; ++i) {
    cin >> y[i];
  }
  vector<int> orig_x = x, orig_y = y;
  vector<pair<int, int> > x_log = gauss(x);
  vector<pair<int, int> > y_log = gauss(y);
  vector<pair<int, int> > merge_log;
  for (int i = 0, j = 0; i < n; ++i) {
    if (y[i] == 0) {
      for (; j < n; ++j) {
        if (x[j] != 0) {
          merge_log.push_back(pair<int, int>(j, j));
          x[j] ^= x[j];
        }
      }
      break;
    }
    while (j < n && highest_bit(x[j]) > highest_bit(y[i])) {
      merge_log.push_back(pair<int, int>(j, j));
      x[j] ^= x[j];
      ++j;
    }
    if (j == n || highest_bit(x[j]) < highest_bit(y[i])) {
      cout << -1 << endl;
      return 0;
    }
    for (int k = j + 1; k < n && x[j] != y[i]; ++k) {
      int b1 = highest_bit(x[k]);
      int b2 = highest_bit(x[j] ^ y[i]);
      if (b2 > b1) {
        break;
      }
      if (b1 == b2) {
        x[j] ^= x[k];
        merge_log.push_back(pair<int, int>(j, k));
      }
    }
    if (x[j] != y[i]) {
      cout << -1 << endl;
      return 0;
    }
    if (j != i) {
      merge_log.push_back(pair<int, int>(j, i));
      x[j] ^= x[i];
      merge_log.push_back(pair<int, int>(i, j));
      x[i] ^= x[j];
      merge_log.push_back(pair<int, int>(j, j));
      x[j] ^= x[j];
    }
    ++j;
  }
  vector<pair<int, int> > result;
  result.insert(result.end(), x_log.begin(), x_log.end());
  result.insert(result.end(), merge_log.begin(), merge_log.end());
  result.insert(result.end(), y_log.rbegin(), y_log.rend());
  x = orig_x;
  y = orig_y;
  cout << result.size() << endl;
  for (int i = 0; i < (int)result.size(); ++i) {
    x[result[i].first] ^= x[result[i].second];
    cout << result[i].first + 1 << " " << result[i].second + 1 << endl;
  }
  if (x != y) {
    return 1;
  }
  return 0;
}

```