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;
const int mod = 1e9 + 7;
inline long long add(long long a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
  return a;
}
inline long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) a += mod;
  return a;
}
inline long long mul(long long a, long long b) {
  return (long long)((long long)a * b % mod);
}
long long n;
vector<long long> a;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t = 1;
  while (t--) {
    cin >> n;
    string s1, s2;
    cin >> s1 >> s2;
    vector<long long> v(2 * n + 1, 0);
    long long c1 = 0, c2 = 0;
    for (long long i = (1); i <= (n); i++) {
      if (s1[i - 1] != s2[i - 1]) {
        s1[i - 1] == '1' ? v[i] = 1 : v[i] = -1;
        s1[i - 1] == '1' ? c1++ : c2++;
      }
    }
    for (long long i = (n + 1); i <= (2 * n); i++) {
      v[i] = v[i - n];
    }
    long long l = 0;
    long long ans = 0;
    for (long long i = (1); i <= (2 * n); i++) {
      l = max(l + v[i], (long long)0);
      ans = max(l, ans);
    }
    if (c1 != c2)
      cout << -1;
    else
      cout << ans;
  }
  return 0;
}
