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 max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
long long fexp(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % 1000000007;
    b /= 2;
    a = a * a % 1000000007;
  }
  return ans;
}
long long inverse(long long a, long long p) { return fexp(a, p - 2); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long T = 1;
  long long t = 0;
  while (t++ < T) {
    long long n;
    string s, t;
    cin >> n;
    cin >> s >> t;
    vector<long long> a(n);
    for (long long i = 0; i < n; i++) {
      if (s[i] == t[i])
        a[i] = 0;
      else if (s[i] == '1')
        a[i] = 1;
      else
        a[i] = -1;
    }
    long long sum = 0;
    for (long long i = 0; i < n; i++) sum += a[i];
    if (sum != 0) {
      cout << -1;
      continue;
    }
    long long maxi = -1;
    sum = 0;
    for (long long i = 0; i < n; i++) {
      sum += a[i];
      if (sum < 0) sum = 0;
      maxi = max(maxi, sum);
    }
    long long mini = 0;
    sum = 0;
    for (long long i = 0; i < n; i++) {
      sum += a[i];
      if (sum > 0) sum = 0;
      mini = min(mini, sum);
    }
    long long ans = max(maxi, -mini);
    cout << ans;
  }
  return 0;
}
