#include <bits/stdc++.h>
using namespace std;
int N, L, R, how[50], x[10009], y[10009];
bool isBase[10009];
vector<pair<int, int> > lft, rgt, mid, ans;
void doSwap(int lin, int i, int j, vector<pair<int, int> > &v) {
  if (i == j) return;
  if (lin == 0)
    swap(x[i], x[j]);
  else
    swap(y[i], y[j]);
  v.push_back({i, j}), v.push_back({j, i}), v.push_back({i, j});
}
int turnToBase(int lin, int x[], vector<pair<int, int> > &ans) {
  for (int i = 1; i <= N; i++) {
    int k = -1;
    for (int j = 0; j < 30; j++)
      if (x[i] & (1 << j)) {
        k = j;
        break;
      }
    if (k == -1) {
      isBase[i] = 0;
      continue;
    }
    isBase[i] = 1;
    for (int j = i + 1; j <= N; j++)
      if (x[j] & (1 << k)) x[j] ^= x[i], ans.push_back({j, i});
  }
  int j = 0;
  for (int i = 1; i <= N; i++)
    if (isBase[i]) doSwap(lin, ++j, i, ans);
  return j;
}
int writeVal(int val) {
  int aux[50];
  for (int i = 1; i <= L; i++) aux[i] = x[i];
  x[0] = val;
  for (int i = 0; i <= L; i++) how[i] = 1 << i;
  for (int i = L; i >= 0; i--) {
    int k = -1;
    for (int j = 0; j < 30; j++)
      if (x[i] & (1 << j)) {
        k = j;
        break;
      }
    if (k == -1) continue;
    for (int j = i - 1; j >= 0; j--)
      if (x[j] & (1 << k)) x[j] ^= x[i], how[j] ^= how[i];
  }
  for (int i = 1; i <= L; i++) x[i] = aux[i];
  if (x[0] != 0) {
    printf("-1\n");
    exit(0);
  }
  return how[0];
}
void doMidStuff() {
  for (int i = 1; i <= R; i++) {
    int msk = writeVal(y[i]);
    for (int j = i; j <= L; j++)
      if (msk & (1 << j)) {
        if (j != i) {
          swap(x[i], x[j]);
          mid.push_back({i, j});
          mid.push_back({j, i});
          mid.push_back({i, j});
        }
        msk ^= 1 << j;
        break;
      }
    for (int j = 1; j <= L; j++)
      if (msk & (1 << j)) x[i] ^= x[j], mid.push_back({i, j});
  }
  for (int i = R + 1; i <= L; i++) mid.push_back({i, i});
}
void check() {
  ifstream f("input");
  f >> N;
  for (int i = 1; i <= N; i++) f >> x[i];
  for (int i = 1; i <= N; i++) f >> y[i];
  f.close();
  ifstream f2("output");
  int L;
  f2 >> L;
  while (L--) {
    int i, j;
    f2 >> i >> j;
    x[i] ^= x[j];
  }
  f2.close();
  for (int i = 1; i <= N; i++) printf("%d%c", x[i], " \n"[i == N]);
  for (int i = 1; i <= N; i++) printf("%d%c", y[i], " \n"[i == N]);
  exit(0);
}
void print(vector<pair<int, int> > ans) {
  printf("%d\n", ans.size());
  for (auto it : ans) printf("%d %d\n", it.first, it.second);
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &x[i]);
  for (int i = 1; i <= N; i++) scanf("%d", &y[i]);
  L = turnToBase(0, x, lft);
  R = turnToBase(1, y, rgt);
  doMidStuff();
  for (auto it : lft) ans.push_back(it);
  for (auto it : mid) ans.push_back(it);
  reverse(rgt.begin(), rgt.end());
  for (auto it : rgt) ans.push_back(it);
  print(ans);
  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, L, R, how[50], x[10009], y[10009];
bool isBase[10009];
vector<pair<int, int> > lft, rgt, mid, ans;
void doSwap(int lin, int i, int j, vector<pair<int, int> > &v) {
  if (i == j) return;
  if (lin == 0)
    swap(x[i], x[j]);
  else
    swap(y[i], y[j]);
  v.push_back({i, j}), v.push_back({j, i}), v.push_back({i, j});
}
int turnToBase(int lin, int x[], vector<pair<int, int> > &ans) {
  for (int i = 1; i <= N; i++) {
    int k = -1;
    for (int j = 0; j < 30; j++)
      if (x[i] & (1 << j)) {
        k = j;
        break;
      }
    if (k == -1) {
      isBase[i] = 0;
      continue;
    }
    isBase[i] = 1;
    for (int j = i + 1; j <= N; j++)
      if (x[j] & (1 << k)) x[j] ^= x[i], ans.push_back({j, i});
  }
  int j = 0;
  for (int i = 1; i <= N; i++)
    if (isBase[i]) doSwap(lin, ++j, i, ans);
  return j;
}
int writeVal(int val) {
  int aux[50];
  for (int i = 1; i <= L; i++) aux[i] = x[i];
  x[0] = val;
  for (int i = 0; i <= L; i++) how[i] = 1 << i;
  for (int i = L; i >= 0; i--) {
    int k = -1;
    for (int j = 0; j < 30; j++)
      if (x[i] & (1 << j)) {
        k = j;
        break;
      }
    if (k == -1) continue;
    for (int j = i - 1; j >= 0; j--)
      if (x[j] & (1 << k)) x[j] ^= x[i], how[j] ^= how[i];
  }
  for (int i = 1; i <= L; i++) x[i] = aux[i];
  if (x[0] != 0) {
    printf("-1\n");
    exit(0);
  }
  return how[0];
}
void doMidStuff() {
  for (int i = 1; i <= R; i++) {
    int msk = writeVal(y[i]);
    for (int j = i; j <= L; j++)
      if (msk & (1 << j)) {
        if (j != i) {
          swap(x[i], x[j]);
          mid.push_back({i, j});
          mid.push_back({j, i});
          mid.push_back({i, j});
        }
        msk ^= 1 << j;
        break;
      }
    for (int j = 1; j <= L; j++)
      if (msk & (1 << j)) x[i] ^= x[j], mid.push_back({i, j});
  }
  for (int i = R + 1; i <= L; i++) mid.push_back({i, i});
}
void check() {
  ifstream f("input");
  f >> N;
  for (int i = 1; i <= N; i++) f >> x[i];
  for (int i = 1; i <= N; i++) f >> y[i];
  f.close();
  ifstream f2("output");
  int L;
  f2 >> L;
  while (L--) {
    int i, j;
    f2 >> i >> j;
    x[i] ^= x[j];
  }
  f2.close();
  for (int i = 1; i <= N; i++) printf("%d%c", x[i], " \n"[i == N]);
  for (int i = 1; i <= N; i++) printf("%d%c", y[i], " \n"[i == N]);
  exit(0);
}
void print(vector<pair<int, int> > ans) {
  printf("%d\n", ans.size());
  for (auto it : ans) printf("%d %d\n", it.first, it.second);
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &x[i]);
  for (int i = 1; i <= N; i++) scanf("%d", &y[i]);
  L = turnToBase(0, x, lft);
  R = turnToBase(1, y, rgt);
  doMidStuff();
  for (auto it : lft) ans.push_back(it);
  for (auto it : mid) ans.push_back(it);
  reverse(rgt.begin(), rgt.end());
  for (auto it : rgt) ans.push_back(it);
  print(ans);
  return 0;
}

```