#include <bits/stdc++.h>
using namespace std;
const int N = 10010;
vector<pair<int, int> > opA, opB;
int a[N];
int b[N];
int n;
void sw(int now, int idx, int a[], vector<pair<int, int> >& op) {
  swap(a[now], a[idx]);
  op.push_back(make_pair(now, idx));
  op.push_back(make_pair(idx, now));
  op.push_back(make_pair(now, idx));
  return;
}
int shift(int x, int s) { return (x >> s) & 1; }
void guass(int a[], vector<pair<int, int> >& op) {
  int now = 0;
  for (int i = 0; i < 31; i++) {
    int idx = -1;
    for (int j = now; j < n; j++) {
      if (shift(a[j], i) == 1) {
        idx = j;
        break;
      }
    }
    if (idx == -1) {
      continue;
    } else {
      if (now != idx) sw(now, idx, a, op);
      for (int j = 0; j < n; j++) {
        if (j != now && shift(a[j], i) == 1) {
          a[j] ^= a[now];
          op.push_back(make_pair(j, now));
        }
      }
      now++;
    }
  }
  return;
}
int next_bit(int x) {
  assert(x != 0);
  int idx = 0;
  while (true) {
    if (x & 1) return idx;
    x >>= 1;
    idx++;
  }
  return idx;
}
void oper(vector<int> p, vector<pair<int, int> >& op, int top) {
  int now = p[0];
  int del = 0;
  for (int i = 1; i < p.size(); i++) {
    int idx = p[i];
    a[now] ^= a[idx];
    op.push_back(make_pair(now, idx));
  }
  for (int i = now; i >= top + 1; i--) sw(i, i - 1, a, op);
  return;
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
    }
    guass(a, opA);
    for (int i = 0; i < n; i++) {
      scanf("%d", &b[i]);
    }
    guass(b, opB);
    vector<pair<int, int> > op;
    int b_m = -1;
    for (int i = 0; i < n; i++) {
      if (b[i] == 0)
        break;
      else {
        b_m = i;
        vector<int> p;
        for (int j = i; j < n; j++) {
          if (b[i] == 0) break;
          if (a[j] == 0) break;
          int nb = next_bit(b[i]), na = next_bit(a[j]);
          if (nb == na) {
            p.push_back(j);
            b[i] ^= a[j];
          } else if (na > nb) {
            puts("-1");
            return 0;
          }
        }
        if (b[i] != 0) {
          puts("-1");
          return 0;
        } else
          oper(p, op, i);
      }
    }
    for (int i = b_m + 1; i < n; i++) {
      if (a[i] == 0) break;
      a[i] ^= a[i];
      op.push_back(make_pair(i, i));
    }
    printf("%d\n", (int)opA.size() + (int)op.size() + (int)opB.size());
    for (int i = 0; i < opA.size(); i++)
      printf("%d %d\n", opA[i].first + 1, opA[i].second + 1);
    for (int i = 0; i < op.size(); i++)
      printf("%d %d\n", op[i].first + 1, op[i].second + 1);
    for (int i = opB.size() - 1; i >= 0; i--)
      printf("%d %d\n", opB[i].first + 1, opB[i].second + 1);
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 = 10010;
vector<pair<int, int> > opA, opB;
int a[N];
int b[N];
int n;
void sw(int now, int idx, int a[], vector<pair<int, int> >& op) {
  swap(a[now], a[idx]);
  op.push_back(make_pair(now, idx));
  op.push_back(make_pair(idx, now));
  op.push_back(make_pair(now, idx));
  return;
}
int shift(int x, int s) { return (x >> s) & 1; }
void guass(int a[], vector<pair<int, int> >& op) {
  int now = 0;
  for (int i = 0; i < 31; i++) {
    int idx = -1;
    for (int j = now; j < n; j++) {
      if (shift(a[j], i) == 1) {
        idx = j;
        break;
      }
    }
    if (idx == -1) {
      continue;
    } else {
      if (now != idx) sw(now, idx, a, op);
      for (int j = 0; j < n; j++) {
        if (j != now && shift(a[j], i) == 1) {
          a[j] ^= a[now];
          op.push_back(make_pair(j, now));
        }
      }
      now++;
    }
  }
  return;
}
int next_bit(int x) {
  assert(x != 0);
  int idx = 0;
  while (true) {
    if (x & 1) return idx;
    x >>= 1;
    idx++;
  }
  return idx;
}
void oper(vector<int> p, vector<pair<int, int> >& op, int top) {
  int now = p[0];
  int del = 0;
  for (int i = 1; i < p.size(); i++) {
    int idx = p[i];
    a[now] ^= a[idx];
    op.push_back(make_pair(now, idx));
  }
  for (int i = now; i >= top + 1; i--) sw(i, i - 1, a, op);
  return;
}
int main() {
  while (~scanf("%d", &n)) {
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
    }
    guass(a, opA);
    for (int i = 0; i < n; i++) {
      scanf("%d", &b[i]);
    }
    guass(b, opB);
    vector<pair<int, int> > op;
    int b_m = -1;
    for (int i = 0; i < n; i++) {
      if (b[i] == 0)
        break;
      else {
        b_m = i;
        vector<int> p;
        for (int j = i; j < n; j++) {
          if (b[i] == 0) break;
          if (a[j] == 0) break;
          int nb = next_bit(b[i]), na = next_bit(a[j]);
          if (nb == na) {
            p.push_back(j);
            b[i] ^= a[j];
          } else if (na > nb) {
            puts("-1");
            return 0;
          }
        }
        if (b[i] != 0) {
          puts("-1");
          return 0;
        } else
          oper(p, op, i);
      }
    }
    for (int i = b_m + 1; i < n; i++) {
      if (a[i] == 0) break;
      a[i] ^= a[i];
      op.push_back(make_pair(i, i));
    }
    printf("%d\n", (int)opA.size() + (int)op.size() + (int)opB.size());
    for (int i = 0; i < opA.size(); i++)
      printf("%d %d\n", opA[i].first + 1, opA[i].second + 1);
    for (int i = 0; i < op.size(); i++)
      printf("%d %d\n", op[i].first + 1, op[i].second + 1);
    for (int i = opB.size() - 1; i >= 0; i--)
      printf("%d %d\n", opB[i].first + 1, opB[i].second + 1);
  }
}

```