#include <bits/stdc++.h>
using namespace std;
const int MAXN = 12000;
int n;
vector<pair<int, int> > vvx;
vector<pair<int, int> > vvy;
int x[MAXN];
int y[MAXN];
void swx(int a, int b) {
  if (a == b) return;
  swap(x[a], x[b]);
  vvx.push_back(make_pair(a, b));
  vvx.push_back(make_pair(b, a));
  vvx.push_back(make_pair(a, b));
}
void swy(int a, int b) {
  if (a == b) return;
  swap(y[a], y[b]);
  vvy.push_back(make_pair(a, b));
  vvy.push_back(make_pair(b, a));
  vvy.push_back(make_pair(a, b));
}
int gg(int x) {
  for (int i = 0; i < 30; ++i)
    if ((x >> i) & 1) return i;
  return -1;
}
int main() {
  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);
  int stx = 0;
  for (int i = 0; i < 30; ++i) {
    int fl = 0;
    for (int j = stx; j < n; ++j) {
      if ((x[j] >> i) & 1) {
        swx(stx, j);
        fl = 1;
        break;
      }
    }
    if (!fl) continue;
    for (int j = 0; j < n; ++j)
      if (j != stx && ((x[j] >> i) & 1))
        vvx.push_back(make_pair(j, stx)), x[j] ^= x[stx];
    ++stx;
  }
  int sty = 0;
  for (int i = 0; i < 30; ++i) {
    int fl = 0;
    for (int j = sty; j < n; ++j) {
      if ((y[j] >> i) & 1) {
        swy(sty, j);
        fl = 1;
        break;
      }
    }
    if (!fl) continue;
    for (int j = 0; j < n; ++j)
      if (j != sty && ((y[j] >> i) & 1))
        vvy.push_back(make_pair(j, sty)), y[j] ^= y[sty];
    ++sty;
  }
  for (int i = 0; i < stx; ++i) {
    for (int j = 0; j < sty; ++j) {
      if (gg(x[i]) == gg(y[j])) {
        for (int k = 0; k < 30; ++k) {
          if (((x[i] >> k) & 1) != ((y[j] >> k) & 1)) {
            for (int l = i + 1; l < stx; ++l)
              if (gg(x[l]) == k) {
                vvx.push_back(make_pair(i, l));
                x[i] ^= x[l];
                break;
              }
          }
        }
      }
    }
  }
  set<int> ssx;
  set<int> ssy;
  for (int i = 0; i < stx; ++i) ssx.insert(x[i]);
  for (int i = 0; i < sty; ++i) ssy.insert(y[i]);
  for (int i = 0; i < sty; ++i) {
    if (!ssx.count(y[i])) {
      cout << -1 << "\n";
      return 0;
    }
  }
  for (int j = 0; j < stx; ++j) {
    if (!ssy.count(x[j])) {
      vvx.push_back(make_pair(j, j));
      x[j] = 0;
      for (int k = j + 1; k < stx; ++k)
        vvx.push_back(make_pair(k - 1, k)), x[k - 1] ^= x[k],
            vvx.push_back(make_pair(k, k)), x[k] = 0;
      --stx;
      j -= 1;
    }
  }
  for (int i = 0; i < n; ++i) assert(x[i] == y[i]);
  while (!vvy.empty()) vvx.push_back(vvy.back()), vvy.pop_back();
  printf("%d\n", (int)vvx.size());
  for (pair<int, int> i : vvx) printf("%d %d\n", i.first + 1, i.second + 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;
const int MAXN = 12000;
int n;
vector<pair<int, int> > vvx;
vector<pair<int, int> > vvy;
int x[MAXN];
int y[MAXN];
void swx(int a, int b) {
  if (a == b) return;
  swap(x[a], x[b]);
  vvx.push_back(make_pair(a, b));
  vvx.push_back(make_pair(b, a));
  vvx.push_back(make_pair(a, b));
}
void swy(int a, int b) {
  if (a == b) return;
  swap(y[a], y[b]);
  vvy.push_back(make_pair(a, b));
  vvy.push_back(make_pair(b, a));
  vvy.push_back(make_pair(a, b));
}
int gg(int x) {
  for (int i = 0; i < 30; ++i)
    if ((x >> i) & 1) return i;
  return -1;
}
int main() {
  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);
  int stx = 0;
  for (int i = 0; i < 30; ++i) {
    int fl = 0;
    for (int j = stx; j < n; ++j) {
      if ((x[j] >> i) & 1) {
        swx(stx, j);
        fl = 1;
        break;
      }
    }
    if (!fl) continue;
    for (int j = 0; j < n; ++j)
      if (j != stx && ((x[j] >> i) & 1))
        vvx.push_back(make_pair(j, stx)), x[j] ^= x[stx];
    ++stx;
  }
  int sty = 0;
  for (int i = 0; i < 30; ++i) {
    int fl = 0;
    for (int j = sty; j < n; ++j) {
      if ((y[j] >> i) & 1) {
        swy(sty, j);
        fl = 1;
        break;
      }
    }
    if (!fl) continue;
    for (int j = 0; j < n; ++j)
      if (j != sty && ((y[j] >> i) & 1))
        vvy.push_back(make_pair(j, sty)), y[j] ^= y[sty];
    ++sty;
  }
  for (int i = 0; i < stx; ++i) {
    for (int j = 0; j < sty; ++j) {
      if (gg(x[i]) == gg(y[j])) {
        for (int k = 0; k < 30; ++k) {
          if (((x[i] >> k) & 1) != ((y[j] >> k) & 1)) {
            for (int l = i + 1; l < stx; ++l)
              if (gg(x[l]) == k) {
                vvx.push_back(make_pair(i, l));
                x[i] ^= x[l];
                break;
              }
          }
        }
      }
    }
  }
  set<int> ssx;
  set<int> ssy;
  for (int i = 0; i < stx; ++i) ssx.insert(x[i]);
  for (int i = 0; i < sty; ++i) ssy.insert(y[i]);
  for (int i = 0; i < sty; ++i) {
    if (!ssx.count(y[i])) {
      cout << -1 << "\n";
      return 0;
    }
  }
  for (int j = 0; j < stx; ++j) {
    if (!ssy.count(x[j])) {
      vvx.push_back(make_pair(j, j));
      x[j] = 0;
      for (int k = j + 1; k < stx; ++k)
        vvx.push_back(make_pair(k - 1, k)), x[k - 1] ^= x[k],
            vvx.push_back(make_pair(k, k)), x[k] = 0;
      --stx;
      j -= 1;
    }
  }
  for (int i = 0; i < n; ++i) assert(x[i] == y[i]);
  while (!vvy.empty()) vvx.push_back(vvy.back()), vvy.pop_back();
  printf("%d\n", (int)vvx.size());
  for (pair<int, int> i : vvx) printf("%d %d\n", i.first + 1, i.second + 1);
  return 0;
}

```