#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
int x[N], y[N];
vector<pair<int, int> > res;
vector<pair<int, int> > tmp;
int stx[N], sty[N];
int MAIN() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &x[i]);
  for (int i = 0; i < n; ++i) scanf("%d", &y[i]);
  for (int i = 0; i < n; ++i) {
    int k = i;
    for (int j = i + 1; j < n; ++j) {
      if (x[j] > x[k]) {
        k = j;
      }
    }
    if (x[k] == 0) break;
    if (k != i) {
      swap(x[i], x[k]);
      res.push_back(make_pair(i, k));
      res.push_back(make_pair(k, i));
      res.push_back(make_pair(i, k));
    }
    for (int j = 1 << 30; j; j >>= 1) {
      if (x[i] & j) {
        stx[i] = j;
        break;
      }
    }
    for (int j = 0; j < n; ++j) {
      if (i == j) continue;
      if (x[j] & stx[i]) {
        x[j] ^= x[i];
        res.push_back(make_pair(j, i));
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    int k = i;
    for (int j = i + 1; j < n; ++j) {
      if (y[j] > y[k]) {
        k = j;
      }
    }
    if (y[k] == 0) break;
    if (i != k) {
      swap(y[i], y[k]);
      tmp.push_back(make_pair(i, k));
      tmp.push_back(make_pair(k, i));
      tmp.push_back(make_pair(i, k));
    }
    for (int j = 1 << 30; j > 0; j >>= 1) {
      if (y[i] & j) {
        sty[i] = j;
        break;
      }
    }
    for (int j = 0; j < n; ++j) {
      if (i == j) continue;
      if (y[j] & sty[i]) {
        y[j] ^= y[i];
        tmp.push_back(make_pair(j, i));
      }
    }
  }
  for (int i = 0; i < n; ++i)
    if (x[i] != y[i]) {
      if (stx[i] < sty[i]) {
        printf("-1\n");
        return 0;
      } else if (stx[i] > sty[i]) {
        x[i] = 0;
        stx[i] = 0;
        res.push_back(make_pair(i, i));
        for (int j = i + 1; j < n; ++j) {
          if (x[j] == 0) break;
          swap(x[j - 1], x[j]);
          swap(stx[j - 1], stx[j]);
          res.push_back(make_pair(j - 1, j));
          res.push_back(make_pair(j, j - 1));
          res.push_back(make_pair(j - 1, j));
        }
        i--;
      } else {
        for (int j = i + 1; j < n; ++j) {
          if (x[j] == 0) break;
          if (y[i] & stx[j]) {
            x[i] ^= x[j];
            res.push_back(make_pair(i, j));
          }
        }
        if (x[i] != y[i]) {
          printf("-1\n");
          return 0;
        }
      }
    }
  reverse(tmp.begin(), tmp.end());
  for (int i = 0; i < tmp.size(); ++i) {
    res.push_back(tmp[i]);
  }
  int size = (int)res.size();
  printf("%d\n", size);
  for (int i = 0; i < size; ++i) {
    printf("%d %d\n", res[i].first + 1, res[i].second + 1);
  }
  return 0;
}
int main() {
  ios ::sync_with_stdio(false);
  cout << fixed << setprecision(16);
  return MAIN();
}

### 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;
const int N = 100005;
int x[N], y[N];
vector<pair<int, int> > res;
vector<pair<int, int> > tmp;
int stx[N], sty[N];
int MAIN() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &x[i]);
  for (int i = 0; i < n; ++i) scanf("%d", &y[i]);
  for (int i = 0; i < n; ++i) {
    int k = i;
    for (int j = i + 1; j < n; ++j) {
      if (x[j] > x[k]) {
        k = j;
      }
    }
    if (x[k] == 0) break;
    if (k != i) {
      swap(x[i], x[k]);
      res.push_back(make_pair(i, k));
      res.push_back(make_pair(k, i));
      res.push_back(make_pair(i, k));
    }
    for (int j = 1 << 30; j; j >>= 1) {
      if (x[i] & j) {
        stx[i] = j;
        break;
      }
    }
    for (int j = 0; j < n; ++j) {
      if (i == j) continue;
      if (x[j] & stx[i]) {
        x[j] ^= x[i];
        res.push_back(make_pair(j, i));
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    int k = i;
    for (int j = i + 1; j < n; ++j) {
      if (y[j] > y[k]) {
        k = j;
      }
    }
    if (y[k] == 0) break;
    if (i != k) {
      swap(y[i], y[k]);
      tmp.push_back(make_pair(i, k));
      tmp.push_back(make_pair(k, i));
      tmp.push_back(make_pair(i, k));
    }
    for (int j = 1 << 30; j > 0; j >>= 1) {
      if (y[i] & j) {
        sty[i] = j;
        break;
      }
    }
    for (int j = 0; j < n; ++j) {
      if (i == j) continue;
      if (y[j] & sty[i]) {
        y[j] ^= y[i];
        tmp.push_back(make_pair(j, i));
      }
    }
  }
  for (int i = 0; i < n; ++i)
    if (x[i] != y[i]) {
      if (stx[i] < sty[i]) {
        printf("-1\n");
        return 0;
      } else if (stx[i] > sty[i]) {
        x[i] = 0;
        stx[i] = 0;
        res.push_back(make_pair(i, i));
        for (int j = i + 1; j < n; ++j) {
          if (x[j] == 0) break;
          swap(x[j - 1], x[j]);
          swap(stx[j - 1], stx[j]);
          res.push_back(make_pair(j - 1, j));
          res.push_back(make_pair(j, j - 1));
          res.push_back(make_pair(j - 1, j));
        }
        i--;
      } else {
        for (int j = i + 1; j < n; ++j) {
          if (x[j] == 0) break;
          if (y[i] & stx[j]) {
            x[i] ^= x[j];
            res.push_back(make_pair(i, j));
          }
        }
        if (x[i] != y[i]) {
          printf("-1\n");
          return 0;
        }
      }
    }
  reverse(tmp.begin(), tmp.end());
  for (int i = 0; i < tmp.size(); ++i) {
    res.push_back(tmp[i]);
  }
  int size = (int)res.size();
  printf("%d\n", size);
  for (int i = 0; i < size; ++i) {
    printf("%d %d\n", res[i].first + 1, res[i].second + 1);
  }
  return 0;
}
int main() {
  ios ::sync_with_stdio(false);
  cout << fixed << setprecision(16);
  return MAIN();
}

```