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 = 1e6 + 15;
vector<long long> A(N, 0);
long long n;
long long kadane(long long x) {
  long long maxi = 0, cur = 0;
  for (long long i = 0; i < n; i++) {
    cur += x * A[i];
    maxi = max(maxi, cur);
    if (cur < 0) cur = 0;
  }
  return maxi;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  string s, t;
  cin >> s >> t;
  long long s0 = 0, s1 = 0, t0 = 0, t1 = 0;
  string str = "";
  for (long long i = 0; i < n; i++) {
    if (s[i] != t[i]) {
      if (s[i] == '1')
        A[i] = -1;
      else
        A[i] = 1;
      str += s[i];
    }
    if (s[i] == '0') s0++;
    if (s[i] == '1') s1++;
    if (t[i] == '0') t0++;
    if (t[i] == '1') t1++;
  }
  if (t0 != s0)
    cout << -1;
  else {
    cout << max(kadane(1), kadane(-1));
  }
  return 0;
}
