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;
pair<int, int> Rv(pair<int, int> a) { return {a.second, a.first}; }
int n, xy, x;
string s, t, a;
set<pair<int, int> > se, es;
void A(int v, int i) {
  se.insert({i, v});
  es.insert({v, i});
}
void A(pair<int, int> a) { A(a.first, a.second); }
void R(int v, int i) {
  se.erase(se.find({i, v}));
  es.erase(es.find({v, i}));
}
void R(pair<int, int> a) { R(a.first, a.second); }
pair<int, int> seb() { return Rv(*se.begin()); }
pair<int, int> esb() { return (*es.begin()); }
pair<int, int> see() { return Rv(*(--se.end())); }
pair<int, int> ese() { return (*(--es.end())); }
void G(pair<int, int> a, pair<int, int> b) {
  R(a), R(b);
  A(a.first + b.first - x, a.second);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> s >> t;
  for (int i = 0; i < n; ++i) {
    if (s[i] != t[i]) {
      a += s[i];
      if (s[i] == '1')
        ++xy;
      else
        --xy;
    }
  }
  if (xy) {
    cout << -1 << endl;
    return 0;
  } else if (a.size() == 0) {
    cout << 0 << endl;
    return 0;
  }
  n = a.size();
  int g = 1, gg = 1;
  for (int i = 1; i < n; ++i) {
    if (a[i - 1] != a[i]) {
      A(g, gg);
      ++gg;
      g = 1;
    } else {
      ++g;
    }
  }
  A(g, gg);
  while (es.size()) {
    if (((int)se.size() & 1) && (se.size() != 1)) G(seb(), see());
    g = esb().first;
    gg = esb().second;
    x = g;
    set<pair<int, int> >::iterator it = se.find({gg, g}), ti = it;
    if (it != se.begin() && it != (--se.end())) {
      ++it, --ti;
      G(Rv(*ti), Rv(*it));
    }
    R({g, gg});
  }
  cout << x << endl;
  return 0;
}
