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;
long long n;
string S, T;
set<long long> st[2];
long long solve() {
  long long ans = 0;
  while (!st[0].empty() && !st[1].empty()) {
    long long sta = *st[0].begin();
    long long tick = 0;
    long long root = 0;
    ans++;
    long long cnt = 0;
    if (*st[1].begin() < sta) {
      sta = *st[1].begin();
      tick = root = 1;
    }
    while (true) {
      st[tick].erase(sta);
      cnt++;
      tick ^= 1;
      auto iter = st[tick].upper_bound(sta);
      if (iter == st[tick].end()) {
        if (tick == root)
          break;
        else {
          if (cnt == 1) return -1;
          st[tick ^ 1].insert(sta);
          break;
        }
      } else
        sta = *iter;
    }
  }
  if (!st[0].empty() || !st[1].empty()) return -1;
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> S >> T;
  for (long long i = (long long)0; i <= (long long)n - 1; i++) {
    if (S[i] == T[i]) continue;
    if (S[i] == '0')
      st[0].insert(i);
    else
      st[1].insert(i);
  }
  cout << solve() << '\n';
}
