#include <bits/stdc++.h>
using namespace std;
const int N = 100000 + 10;
const int MOD = 1e9 + 7;
int n;
long long ox[N], oy[N];
long long x[N], y[N];
int tox[N];
long long c[N], d[N];
int fx[40], fy[40];
vector<pair<int, int> > vec0;
vector<pair<int, int> > vec1;
long long readInt() {
  int x;
  scanf("%d", &x);
  return x;
}
void solve() {
  cin >> n;
  vec0.clear();
  vec1.clear();
  memset(fx, -1, sizeof fx);
  memset(tox, -1, sizeof tox);
  for (int i = 0; i < n; ++i) {
    x[i] = readInt();
    c[i] = x[i];
    for (int j = 31; j >= 0; --j) {
      if (x[i] >> j & 1) {
        if (fx[j] >= 0) {
          x[i] ^= x[fx[j]];
          vec0.push_back(make_pair(i, fx[j]));
        } else {
          fx[j] = i;
          tox[i] = j;
          break;
        }
      } else {
      }
    }
  }
  memset(fy, -1, sizeof fy);
  for (int i = 0; i < n; ++i) {
    y[i] = readInt();
    d[i] = y[i];
    for (int j = 31; j >= 0; --j) {
      if (y[i] >> j & 1) {
        if (fy[j] >= 0) {
          y[i] ^= y[fy[j]];
          vec1.push_back(make_pair(i, fy[j]));
        } else {
          fy[j] = i;
          break;
        }
      } else {
      }
    }
  }
  for (int j = 31; j >= 0; --j) {
    if (fy[j] >= 0) {
      if (fx[j] < 0) {
        cout << "-1" << endl;
        return;
      }
      if (fx[j] != fy[j]) {
        swap(x[fx[j]], x[fy[j]]);
        swap(tox[fx[j]], tox[fy[j]]);
        vec0.push_back(make_pair(fx[j], fy[j]));
        vec0.push_back(make_pair(fy[j], fx[j]));
        vec0.push_back(make_pair(fx[j], fy[j]));
        if (tox[fx[j]] >= 0) {
          fx[tox[fx[j]]] = fx[j];
        }
        if (tox[fy[j]] >= 0) {
          fx[tox[fy[j]]] = fy[j];
        }
      }
      for (int k = j - 1; k >= 0; --k) {
        if ((x[fx[j]] >> k & 1) ^ (y[fx[j]] >> k & 1)) {
          if (fx[k] < 0) {
            cout << "-1" << endl;
            return;
          }
          x[fx[j]] ^= x[fx[k]];
          vec0.push_back(make_pair(fx[j], fx[k]));
        }
      }
    } else {
      if (fx[j] >= 0) {
        x[fx[j]] = 0;
        vec0.push_back(make_pair(fx[j], fx[j]));
      }
    }
  }
  for (int i = (int)vec1.size() - 1; i >= 0; --i) {
    vec0.push_back(vec1[i]);
  }
  cout << vec0.size() << endl;
  for (int i = 0; i < vec0.size(); ++i) {
    printf("%d %d\n", vec0[i].first + 1, vec0[i].second + 1);
    c[vec0[i].first] ^= c[vec0[i].second];
  }
  for (int i = 0; i < n; ++i) {
    if (c[i] != d[i]) {
      cout << "Wrong!" << endl;
      for (int j = 0; j < n; ++j) {
        cout << ox[j] << ' ';
      }
      cout << endl;
      for (int j = 0; j < n; ++j) {
        cout << oy[j] << ' ';
      }
      cout << endl;
      assert(0);
    }
  }
}
int main() {
  solve();
  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 N = 100000 + 10;
const int MOD = 1e9 + 7;
int n;
long long ox[N], oy[N];
long long x[N], y[N];
int tox[N];
long long c[N], d[N];
int fx[40], fy[40];
vector<pair<int, int> > vec0;
vector<pair<int, int> > vec1;
long long readInt() {
  int x;
  scanf("%d", &x);
  return x;
}
void solve() {
  cin >> n;
  vec0.clear();
  vec1.clear();
  memset(fx, -1, sizeof fx);
  memset(tox, -1, sizeof tox);
  for (int i = 0; i < n; ++i) {
    x[i] = readInt();
    c[i] = x[i];
    for (int j = 31; j >= 0; --j) {
      if (x[i] >> j & 1) {
        if (fx[j] >= 0) {
          x[i] ^= x[fx[j]];
          vec0.push_back(make_pair(i, fx[j]));
        } else {
          fx[j] = i;
          tox[i] = j;
          break;
        }
      } else {
      }
    }
  }
  memset(fy, -1, sizeof fy);
  for (int i = 0; i < n; ++i) {
    y[i] = readInt();
    d[i] = y[i];
    for (int j = 31; j >= 0; --j) {
      if (y[i] >> j & 1) {
        if (fy[j] >= 0) {
          y[i] ^= y[fy[j]];
          vec1.push_back(make_pair(i, fy[j]));
        } else {
          fy[j] = i;
          break;
        }
      } else {
      }
    }
  }
  for (int j = 31; j >= 0; --j) {
    if (fy[j] >= 0) {
      if (fx[j] < 0) {
        cout << "-1" << endl;
        return;
      }
      if (fx[j] != fy[j]) {
        swap(x[fx[j]], x[fy[j]]);
        swap(tox[fx[j]], tox[fy[j]]);
        vec0.push_back(make_pair(fx[j], fy[j]));
        vec0.push_back(make_pair(fy[j], fx[j]));
        vec0.push_back(make_pair(fx[j], fy[j]));
        if (tox[fx[j]] >= 0) {
          fx[tox[fx[j]]] = fx[j];
        }
        if (tox[fy[j]] >= 0) {
          fx[tox[fy[j]]] = fy[j];
        }
      }
      for (int k = j - 1; k >= 0; --k) {
        if ((x[fx[j]] >> k & 1) ^ (y[fx[j]] >> k & 1)) {
          if (fx[k] < 0) {
            cout << "-1" << endl;
            return;
          }
          x[fx[j]] ^= x[fx[k]];
          vec0.push_back(make_pair(fx[j], fx[k]));
        }
      }
    } else {
      if (fx[j] >= 0) {
        x[fx[j]] = 0;
        vec0.push_back(make_pair(fx[j], fx[j]));
      }
    }
  }
  for (int i = (int)vec1.size() - 1; i >= 0; --i) {
    vec0.push_back(vec1[i]);
  }
  cout << vec0.size() << endl;
  for (int i = 0; i < vec0.size(); ++i) {
    printf("%d %d\n", vec0[i].first + 1, vec0[i].second + 1);
    c[vec0[i].first] ^= c[vec0[i].second];
  }
  for (int i = 0; i < n; ++i) {
    if (c[i] != d[i]) {
      cout << "Wrong!" << endl;
      for (int j = 0; j < n; ++j) {
        cout << ox[j] << ' ';
      }
      cout << endl;
      for (int j = 0; j < n; ++j) {
        cout << oy[j] << ' ';
      }
      cout << endl;
      assert(0);
    }
  }
}
int main() {
  solve();
  return 0;
}

```