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;
string a, b;
set<int> s, t;
int ans = 0;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  int n;
  cin >> n;
  cin >> a >> b;
  for (int i = 0; i < n; i++) {
    if (a[i] == b[i]) continue;
    if (a[i] == '0' && b[i] == '1')
      s.insert(i);
    else
      t.insert(i);
  }
  if (t.size() != s.size()) {
    cout << "-1\n";
    return 0;
  }
  while (!s.empty() && !t.empty()) {
    int L = -1, ok = 0;
    if (*(s.begin()) < *(t.begin())) ok = 1;
    while (true) {
      if (ok) {
        auto it = s.lower_bound(L);
        if (it == s.end()) break;
        auto nxt = t.lower_bound(*it);
        if (nxt == t.end()) break;
        L = int(*nxt);
        s.erase(it);
        t.erase(nxt);
      } else {
        auto it = t.lower_bound(L);
        if (it == t.end()) break;
        auto nxt = s.lower_bound(*it);
        if (nxt == s.end()) break;
        L = *nxt;
        t.erase(it);
        s.erase(nxt);
      }
    }
    ans++;
  }
  cout << ans << "\n";
}
