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;
long long n;
string s, t;
map<long long, long long> mp;
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  cin >> s >> t;
  string ss = s, tt = t;
  sort(ss.begin(), ss.end());
  sort(tt.begin(), tt.end());
  if (ss != tt) {
    cout << -1 << endl;
    return 0;
  }
  string tmp;
  for (int i = 0; i < n; i++) {
    if (s[i] != t[i]) tmp += s[i];
  }
  s = tmp;
  if (s.size() == 0) {
    cout << 0 << endl;
    return 0;
  }
  long long cnt = 1, id = 0;
  for (int i = 1; i < s.size(); i++) {
    if (s[i] == s[i - 1])
      cnt++;
    else {
      mp[id++] = cnt;
      cnt = 1;
    }
  }
  if (cnt > 0) mp[id] = cnt;
  long long ans = 0;
  while (mp.size()) {
    ans++;
    for (auto it = mp.begin(); it != mp.end(); it++) {
      auto suc = it;
      suc++;
      if (suc != mp.end())
        it->second--;
      else {
        if (mp.size() % 2 == 0) it->second--;
      }
    }
    for (auto it = mp.begin(); it != mp.end();) {
      if (it->second == 0) {
        auto cur = it;
        auto suc = it;
        suc++;
        if (it != mp.begin() && suc != mp.end()) {
          it--;
          it->second += suc->second;
          mp.erase(suc);
          mp.erase(cur);
          it++;
        } else {
          it++;
          mp.erase(cur);
        }
      } else
        it++;
    }
  }
  cout << ans << endl;
  return 0;
}
