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(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  string s, t;
  cin >> s >> t;
  set<long long int> a;
  set<long long int> b;
  long long int o = 0;
  long long int z = 0;
  for (long long int i = 0; i < n; i++) {
    if (s[i] != t[i]) {
      if (s[i] == '1') {
        a.insert(i);
        o++;
      } else {
        b.insert(i);
        z++;
      }
    }
  }
  if (z != o) {
    cout << -1;
  } else {
    long long int cnt = 0;
    long long int ans = 0;
    auto it1 = a.begin();
    auto it2 = b.begin();
    while (!a.empty()) {
      if ((*it1) < (*it2)) {
        cnt++;
        a.erase(it1);
        it1 = a.upper_bound((*it2));
      } else {
        cnt++;
        b.erase(it2);
        it2 = b.upper_bound((*it1));
      }
      if (it1 == a.end() || it2 == b.end()) {
        ans++;
        if (cnt % 2 == 1) {
          if (it1 != a.end()) {
            a.erase(it1);
          } else {
            b.erase(it2);
          }
        }
        cnt = 0;
        it1 = a.begin();
        it2 = b.begin();
      }
    }
    cout << ans;
  }
  return 0;
}
