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 N = 1e6 + 6;
int a[N];
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, cnt = 0;
  cin >> n;
  string S, T, s, t;
  cin >> S >> T;
  for (int i = 0; i < n; i++) {
    if (S[i] != T[i]) s += S[i], t += T[i];
    if (S[i] == '0') cnt++;
    if (T[i] == '0') cnt--;
  }
  if (cnt != 0) return cout << -1 << "\n", 0;
  n = s.size();
  int m = 1, block = 1;
  char cur = s[0];
  for (int i = 1; i < n; i++) {
    if (s[i] != cur)
      a[m++] = block, block = 1, cur = s[i];
    else
      block++;
  }
  a[m] = block;
  if (s[0] == s[n - 1] && block != n) a[1] += a[m], m--;
  int mn = 0, mx = 0, MN = 0, MX = 0, sum = 0;
  for (int i = 1; i <= m; i++) {
    sum += a[i] * (i % 2 ? 1 : -1);
    MX = max(MX, sum - mn), MN = min(MN, sum - mx);
    mx = max(mx, sum), mn = min(mn, sum);
  }
  int ans = max(max(MX, -MN), max(sum - MN, MX - sum));
  cout << ans << "\n";
  return 0;
}
