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;
int n, ans;
string S, T;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  set<int> s[2];
  cin >> n >> S >> T;
  for (int i = 0; i < n; ++i) {
    if (S[i] == '1' && T[i] == '0')
      s[1].insert(i);
    else if (S[i] == '0' && T[i] == '1')
      s[0].insert(i);
  }
  if (s[1].size() != s[0].size()) {
    cout << -1;
    return 0;
  }
  while (1) {
    int curr = -1, ok = 0;
    if (s[0].empty() && s[1].empty()) break;
    if (s[0].empty() || s[1].empty()) {
      ans += max(s[0].size(), s[1].size());
      break;
    }
    ++ans;
    if (*s[1].begin() < *s[0].begin()) ok ^= 1;
    while (1) {
      auto it = s[ok].lower_bound(curr);
      if (it == s[ok].end())
        break;
      else {
        curr = *it;
        s[ok].erase(it);
        ok ^= 1;
      }
    }
  }
  cout << ans;
}
