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;
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) int idx,
               __attribute__((unused)) int LINE_NUM) {
  cerr << endl;
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, int LINE_NUM, Head H, Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
template <class T>
bool ckmin(T& a, T& b) {
  bool r = a > b;
  a = min(a, b);
  return r;
}
template <class T>
bool ckmax(T& a, T& b) {
  bool r = a < b;
  a = max(a, b);
  return r;
}
const int MOD = 1e9 + 7;
const long long INF = 1e18;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  string second, t;
  cin >> second >> t;
  int cnts = 0;
  int cntt = 0;
  for (int i = 0; i < n; i++) {
    if (second[i] == '1') cnts++;
    if (t[i] == '1') cntt++;
  }
  if (cnts != cntt) {
    cout << -1 << '\n';
    ;
    return 0;
  }
  int currP = 0;
  int currN = 0;
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (second[i] != t[i]) {
      if (second[i] == '1') {
        currP = max(1, currP + 1);
        currN = min(1, currN + 1);
      } else {
        currP = max(-1, currP - 1);
        currN = min(-1, currN - 1);
      }
    }
    ans = max(ans, currP);
    ans = max(ans, -currN);
  }
  cout << ans << '\n';
  ;
}
