#include <bits/stdc++.h>
using namespace std;
struct XorGauss {
  static const int BIT = 32, maxn = 10005;
  bitset<BIT> u[maxn];
  int pos[maxn];
  virtual int gauss(int n, int m) {
    int r = 0;
    for (int j = 0, k; j < m; ++j) {
      for (k = r; k < n && !u[k][j]; ++k)
        ;
      if (k >= n) continue;
      if (k != r) {
        swap(u[r], u[k]);
      }
      pos[r] = j;
      for (int i = 0; i < n; ++i)
        if (i != r && u[i][j]) {
          u[i] ^= u[r];
        }
      ++r;
    }
    return r;
  }
};
struct XorGaussP : public XorGauss {
  virtual int gauss(int n, int m, vector<pair<int, int> > &op) {
    int r = 0;
    for (int j = 0, k; j < m; ++j) {
      for (k = r; k < n && !u[k][j]; ++k)
        ;
      if (k >= n) continue;
      if (k != r) {
        op.push_back(make_pair((k), (r)));
        op.push_back(make_pair((r), (k)));
        op.push_back(make_pair((k), (r)));
        swap(u[r], u[k]);
      }
      pos[r] = j;
      for (int i = 0; i < n; ++i)
        if (i != r && u[i][j]) {
          u[i] ^= u[r];
          op.push_back(make_pair((i), (r)));
        }
      ++r;
    }
    return r;
  }
} svx, svy;
int main(int argc, char const *argv[]) {
  int n;
  scanf("%d", &n);
  static int X[10005], Y[10005];
  for (int(i) = 0; (i) < (int)(n); ++(i)) scanf("%d", X + i), svx.u[i] = X[i];
  for (int(i) = 0; (i) < (int)(n); ++(i)) scanf("%d", Y + i), svy.u[i] = Y[i];
  vector<pair<int, int> > opx, opy;
  int rx = svx.gauss(n, 32, opx), ry = svy.gauss(n, 32, opy);
  if (rx < ry) return puts("-1"), 0;
  for (int(i) = 0; (i) < (int)(ry); ++(i)) {
    for (int(j) = 0; (j) < (int)(rx); ++(j))
      if (svx.u[i][svx.pos[j]] != svy.u[i][svx.pos[j]]) {
        svx.u[i] ^= svx.u[j];
        opx.push_back(make_pair((i), (j)));
      }
    if (svx.u[i] != svy.u[i]) return puts("-1"), 0;
  }
  for (int(i) = (ry); (i) < (int)(rx); ++(i))
    svx.u[i] = 0, opx.push_back(make_pair((i), (i)));
  printf("%d\n", (int)(opx).size() + (int)(opy).size());
  for (auto p : opx) printf("%d %d\n", p.first + 1, p.second + 1);
  for (int(i) = (int)((int)(opy).size() - 1); (i) >= (int)(0); --(i))
    printf("%d %d\n", opy[i].first + 1, opy[i].second + 1);
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
struct XorGauss {
  static const int BIT = 32, maxn = 10005;
  bitset<BIT> u[maxn];
  int pos[maxn];
  virtual int gauss(int n, int m) {
    int r = 0;
    for (int j = 0, k; j < m; ++j) {
      for (k = r; k < n && !u[k][j]; ++k)
        ;
      if (k >= n) continue;
      if (k != r) {
        swap(u[r], u[k]);
      }
      pos[r] = j;
      for (int i = 0; i < n; ++i)
        if (i != r && u[i][j]) {
          u[i] ^= u[r];
        }
      ++r;
    }
    return r;
  }
};
struct XorGaussP : public XorGauss {
  virtual int gauss(int n, int m, vector<pair<int, int> > &op) {
    int r = 0;
    for (int j = 0, k; j < m; ++j) {
      for (k = r; k < n && !u[k][j]; ++k)
        ;
      if (k >= n) continue;
      if (k != r) {
        op.push_back(make_pair((k), (r)));
        op.push_back(make_pair((r), (k)));
        op.push_back(make_pair((k), (r)));
        swap(u[r], u[k]);
      }
      pos[r] = j;
      for (int i = 0; i < n; ++i)
        if (i != r && u[i][j]) {
          u[i] ^= u[r];
          op.push_back(make_pair((i), (r)));
        }
      ++r;
    }
    return r;
  }
} svx, svy;
int main(int argc, char const *argv[]) {
  int n;
  scanf("%d", &n);
  static int X[10005], Y[10005];
  for (int(i) = 0; (i) < (int)(n); ++(i)) scanf("%d", X + i), svx.u[i] = X[i];
  for (int(i) = 0; (i) < (int)(n); ++(i)) scanf("%d", Y + i), svy.u[i] = Y[i];
  vector<pair<int, int> > opx, opy;
  int rx = svx.gauss(n, 32, opx), ry = svy.gauss(n, 32, opy);
  if (rx < ry) return puts("-1"), 0;
  for (int(i) = 0; (i) < (int)(ry); ++(i)) {
    for (int(j) = 0; (j) < (int)(rx); ++(j))
      if (svx.u[i][svx.pos[j]] != svy.u[i][svx.pos[j]]) {
        svx.u[i] ^= svx.u[j];
        opx.push_back(make_pair((i), (j)));
      }
    if (svx.u[i] != svy.u[i]) return puts("-1"), 0;
  }
  for (int(i) = (ry); (i) < (int)(rx); ++(i))
    svx.u[i] = 0, opx.push_back(make_pair((i), (i)));
  printf("%d\n", (int)(opx).size() + (int)(opy).size());
  for (auto p : opx) printf("%d %d\n", p.first + 1, p.second + 1);
  for (int(i) = (int)((int)(opy).size() - 1); (i) >= (int)(0); --(i))
    printf("%d %d\n", opy[i].first + 1, opy[i].second + 1);
  return 0;
}

```