Naman has two binary strings s and t of length n (a binary string is a string which only consists of the characters "0" and "1"). He wants to convert s into t using the following operation as few times as possible.

In one operation, he can choose any subsequence of s and rotate it clockwise once.

For example, if s = 1110100, he can choose a subsequence corresponding to indices (1-based) \{2, 6, 7 \} and rotate them clockwise. The resulting string would then be s = 1010110.

A string a is said to be a subsequence of string b if a can be obtained from b by deleting some characters without changing the ordering of the remaining characters.

To perform a clockwise rotation on a sequence c of size k is to perform an operation which sets c_1:=c_k, c_2:=c_1, c_3:=c_2, …, c_k:=c_{k-1} simultaneously.

Determine the minimum number of operations Naman has to perform to convert s into t or say that it is impossible. 

Input

The first line contains a single integer n (1 ≤ n ≤ 10^6) — the length of the strings.

The second line contains the binary string s of length n.

The third line contains the binary string t of length n.

Output

If it is impossible to convert s to t after any number of operations, print -1.

Otherwise, print the minimum number of operations required.

Examples

Input


6
010000
000001


Output


1

Input


10
1111100000
0000011111


Output


5

Input


8
10101010
01010101


Output


1

Input


10
1111100000
1111100001


Output


-1

Note

In the first test, Naman can choose the subsequence corresponding to indices \{2, 6\} and rotate it once to convert s into t.

In the second test, he can rotate the subsequence corresponding to all indices 5 times. It can be proved, that it is the minimum required number of operations.

In the last test, it is impossible to convert s into t.
#include <bits/stdc++.h>
using namespace std;
const int N = 1e6;
int n;
string a, b;
vector<pair<int, int>> f;
void insert(pair<int, int> u) {
  if (int((f).size()) and f.back().first == u.first) {
    f.back().second += u.second;
  } else if (u.second) {
    f.push_back(u);
  }
}
void fix() {
  if (f.back().first == f[0].first) {
    f[0].second += f.back().second;
    f.pop_back();
  }
}
void solve() {
  fix();
  vector<pair<int, int>> g(f);
  f.clear();
  for (auto &u : g) {
    u.second -= 1, insert(u);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> a >> b;
  int ca = 0, cb = 0;
  for (int i = (0); i < (n); ++i) {
    ca += a[i] - '0';
    cb += b[i] - '0';
    if (a[i] != b[i]) {
      insert(make_pair(a[i] - '0', 1));
    }
  }
  if (ca != cb) {
    cout << "-1\n";
  } else {
    int res = 0;
    while (int((f).size())) {
      solve();
      res += 1;
    }
    cout << res << '\n';
  }
  return 0;
}
