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 i, j, n, m, k;
string S, T;
string s;
void nhap() {
  cin >> n >> S >> T;
  S = " " + S;
  T = " " + T;
}
bool xet(string A, string B) {
  int dem[3];
  dem[0] = dem[1] = 0;
  for (int i = int(1); i <= int(n); i++) dem[A[i] - '0']++;
  for (int i = int(1); i <= int(n); i++) dem[B[i] - '0']--;
  if (!dem[0] && !dem[1]) return true;
  return false;
}
void chuyen(string A, string B) {
  s = " ";
  for (int i = int(1); i <= int(n); i++)
    if (A[i] != B[i]) s += A[i];
}
void xuli() {
  if (!xet(S, T)) {
    cout << -1 << "\n";
    return;
  }
  chuyen(S, T);
  int minn = 0, maxx = 0, ct = 0;
  for (int i = int(1); i <= int(s.size() - 1); i++) {
    if (s[i] == '1')
      ct++;
    else
      ct--;
    minn = min(minn, ct);
    maxx = max(maxx, ct);
  }
  cout << maxx - minn << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  nhap();
  xuli();
}
