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 main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, cur, ans = 0;
  cin >> n;
  string a, b;
  cin >> a;
  cin >> b;
  set<int> s;
  set<int> t;
  vector<int> v;
  vector<int> w;
  vector<int> x;
  vector<int> y;
  set<int>::iterator it1;
  set<int>::iterator it2;
  for (int i = 0; i < n; ++i) {
    if (a[i] == '1' && b[i] == '0')
      s.insert(i);
    else if (a[i] == '0' && b[i] == '1')
      t.insert(i);
  }
  if ((int)(s).size() != (int)(t).size()) {
    cout << -1;
    return 0;
  }
  while ((int)(s).size()) {
    ++ans;
    cur = -1;
    v.clear();
    w.clear();
    x.clear();
    y.clear();
    while (1) {
      it1 = s.lower_bound(cur);
      if (it1 == s.end()) break;
      it2 = t.lower_bound(*it1);
      if (it2 == t.end()) break;
      cur = *it2;
      v.push_back(*it1);
      w.push_back(*it2);
      s.erase(it1);
      t.erase(it2);
    }
    for (int i : v) s.insert(i);
    for (int i : w) t.insert(i);
    cur = -1;
    while (1) {
      it1 = t.lower_bound(cur);
      if (it1 == t.end()) break;
      it2 = s.lower_bound(*it1);
      if (it2 == s.end()) break;
      cur = *it2;
      x.push_back(*it1);
      y.push_back(*it2);
      t.erase(it1);
      s.erase(it2);
    }
    if ((int)(v).size() > (int)(x).size()) {
      for (int i : x) t.insert(i);
      for (int i : y) s.insert(i);
      for (int i : v) s.erase(s.lower_bound(i));
      for (int i : w) t.erase(t.lower_bound(i));
    }
  }
  cout << ans;
  return 0;
}
