#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100000;
int a[MAXN], b[MAXN];
int pa[MAXN], push_back[MAXN];
int u[MAXN];
vector<pair<int, int> > ans, ansb;
void perf(int i, int j) {
  a[i] ^= a[j];
  ans.push_back(make_pair(i, j));
}
void perfb(int i, int j) {
  b[i] ^= b[j];
  ansb.push_back(make_pair(i, j));
}
void swapb(int i, int j) {
  if (i == j) return;
  perfb(i, j);
  perfb(j, i);
  perfb(i, j);
}
int main() {
  ios::sync_with_stdio(false);
  cout.precision(10);
  cout << fixed;
  int N;
  cin >> N;
  for (int i = 0; i < (int)(N); ++i) {
    cin >> a[i];
    pa[i] = a[i];
  }
  for (int i = 0; i < (int)(N); ++i) {
    cin >> b[i];
    push_back[i] = b[i];
  }
  vector<int> p(30);
  for (int k = (int)(30) - 1; k >= 0; --k) {
    int x = -1;
    for (int i = 0; i < (int)(N); ++i) {
      if (((a[i] >> k) & 1) && !u[i]) {
        x = i;
        break;
      }
    }
    p[k] = x;
    if (x == -1) continue;
    u[x] = 1;
    for (int i = 0; i < (int)(N); ++i) {
      if (x != i && ((a[i] >> k) & 1)) {
        perf(i, x);
      }
    }
  }
  bool ok = true;
  for (int i = 0; i < (int)(N); ++i) {
    if (u[i]) continue;
    for (int j = (int)(30) - 1; j >= 0; --j) {
      if (((a[i] ^ b[i]) >> j) & 1) {
        if (p[j] == -1)
          ok = false;
        else
          perf(i, p[j]);
      }
    }
  }
  for (int k = 0; k < (int)(30); ++k) {
    if (p[k] == -1) continue;
    int x = 0;
    for (int i = (int)(30) - 1; i >= 0; --i) {
      if ((b[p[k]] >> i) & 1) {
        if (p[i] == -1) {
          ok = false;
          break;
        }
        b[p[k]] ^= a[p[i]];
        x ^= (1 << i);
      }
    }
    if (!ok) break;
    assert(b[p[k]] == 0);
    b[p[k]] = x;
  }
  if (ok) {
    for (int i = 0; i < (int)(30); ++i) {
      if (p[i] == -1) continue;
      a[p[i]] = 1 << i;
    }
    for (int k = (int)(30) - 1; k >= 0; --k) {
      int x = -1;
      for (int i = (int)(k + 1) - 1; i >= 0; --i) {
        if (p[i] == -1) continue;
        if ((b[p[i]] >> k) & 1) {
          x = p[i];
          break;
        }
      }
      if (x == -1 && p[k] == -1) continue;
      if (x == -1) {
        continue;
      }
      if (p[k] == -1) {
        ok = false;
        break;
      }
      swapb(x, p[k]);
      for (int i = 0; i < (int)(k); ++i) {
        if (p[i] == -1) continue;
        if ((b[p[i]] >> k) & 1) {
          perfb(p[i], p[k]);
        }
      }
    }
    for (int i = 0; i < (int)(30); ++i) {
      if (p[i] == -1) continue;
      if (b[p[i]] == 0) {
        perfb(p[i], p[i]);
        b[p[i]] = 1 << i;
      }
      for (int k = i + 1; k < 30; ++k) {
        if (p[k] == -1) continue;
        if ((b[p[k]] >> i) & 1) perfb(p[k], p[i]);
      }
    }
    for (int i = 0; i < (int)(30); ++i)
      assert(p[i] == -1 || b[p[i]] == (1 << i));
  }
  reverse((ansb).begin(), (ansb).end());
  for (int i = 0; i < (int)(ansb.size()); ++i) ans.push_back(ansb[i]);
  if (!ok) {
    cout << -1 << '\n';
    return 0;
  }
  cout << ans.size() << '\n';
  for (int i = 0; i < (int)(ans.size()); ++i) {
    cout << ans[i].first + 1 << ' ' << ans[i].second + 1 << '\n';
    pa[ans[i].first] ^= pa[ans[i].second];
  }
  for (int i = 0; i < (int)(N); ++i) assert(pa[i] == push_back[i]);
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 = 100000;
int a[MAXN], b[MAXN];
int pa[MAXN], push_back[MAXN];
int u[MAXN];
vector<pair<int, int> > ans, ansb;
void perf(int i, int j) {
  a[i] ^= a[j];
  ans.push_back(make_pair(i, j));
}
void perfb(int i, int j) {
  b[i] ^= b[j];
  ansb.push_back(make_pair(i, j));
}
void swapb(int i, int j) {
  if (i == j) return;
  perfb(i, j);
  perfb(j, i);
  perfb(i, j);
}
int main() {
  ios::sync_with_stdio(false);
  cout.precision(10);
  cout << fixed;
  int N;
  cin >> N;
  for (int i = 0; i < (int)(N); ++i) {
    cin >> a[i];
    pa[i] = a[i];
  }
  for (int i = 0; i < (int)(N); ++i) {
    cin >> b[i];
    push_back[i] = b[i];
  }
  vector<int> p(30);
  for (int k = (int)(30) - 1; k >= 0; --k) {
    int x = -1;
    for (int i = 0; i < (int)(N); ++i) {
      if (((a[i] >> k) & 1) && !u[i]) {
        x = i;
        break;
      }
    }
    p[k] = x;
    if (x == -1) continue;
    u[x] = 1;
    for (int i = 0; i < (int)(N); ++i) {
      if (x != i && ((a[i] >> k) & 1)) {
        perf(i, x);
      }
    }
  }
  bool ok = true;
  for (int i = 0; i < (int)(N); ++i) {
    if (u[i]) continue;
    for (int j = (int)(30) - 1; j >= 0; --j) {
      if (((a[i] ^ b[i]) >> j) & 1) {
        if (p[j] == -1)
          ok = false;
        else
          perf(i, p[j]);
      }
    }
  }
  for (int k = 0; k < (int)(30); ++k) {
    if (p[k] == -1) continue;
    int x = 0;
    for (int i = (int)(30) - 1; i >= 0; --i) {
      if ((b[p[k]] >> i) & 1) {
        if (p[i] == -1) {
          ok = false;
          break;
        }
        b[p[k]] ^= a[p[i]];
        x ^= (1 << i);
      }
    }
    if (!ok) break;
    assert(b[p[k]] == 0);
    b[p[k]] = x;
  }
  if (ok) {
    for (int i = 0; i < (int)(30); ++i) {
      if (p[i] == -1) continue;
      a[p[i]] = 1 << i;
    }
    for (int k = (int)(30) - 1; k >= 0; --k) {
      int x = -1;
      for (int i = (int)(k + 1) - 1; i >= 0; --i) {
        if (p[i] == -1) continue;
        if ((b[p[i]] >> k) & 1) {
          x = p[i];
          break;
        }
      }
      if (x == -1 && p[k] == -1) continue;
      if (x == -1) {
        continue;
      }
      if (p[k] == -1) {
        ok = false;
        break;
      }
      swapb(x, p[k]);
      for (int i = 0; i < (int)(k); ++i) {
        if (p[i] == -1) continue;
        if ((b[p[i]] >> k) & 1) {
          perfb(p[i], p[k]);
        }
      }
    }
    for (int i = 0; i < (int)(30); ++i) {
      if (p[i] == -1) continue;
      if (b[p[i]] == 0) {
        perfb(p[i], p[i]);
        b[p[i]] = 1 << i;
      }
      for (int k = i + 1; k < 30; ++k) {
        if (p[k] == -1) continue;
        if ((b[p[k]] >> i) & 1) perfb(p[k], p[i]);
      }
    }
    for (int i = 0; i < (int)(30); ++i)
      assert(p[i] == -1 || b[p[i]] == (1 << i));
  }
  reverse((ansb).begin(), (ansb).end());
  for (int i = 0; i < (int)(ansb.size()); ++i) ans.push_back(ansb[i]);
  if (!ok) {
    cout << -1 << '\n';
    return 0;
  }
  cout << ans.size() << '\n';
  for (int i = 0; i < (int)(ans.size()); ++i) {
    cout << ans[i].first + 1 << ' ' << ans[i].second + 1 << '\n';
    pa[ans[i].first] ^= pa[ans[i].second];
  }
  for (int i = 0; i < (int)(N); ++i) assert(pa[i] == push_back[i]);
  return 0;
}

```