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 Max = 1e6 + 10;
const int Mod = 1e9 + 7;
const double PI = 3.141592653589793238463;
long long n, cum[Max + 10];
long long cal(long long m) {
  long long ans = 0, i;
  for (i = 1; i <= n; i++) ans += min(cum[i] % m, m - (cum[i] % m));
  return ans;
}
int main() {
  (ios_base::sync_with_stdio(false), cin.tie(NULL));
  ;
  long long i, j, m, p, a, sum = 0, k, t, b, c, d, cnt = 0, q, l, r;
  string str, ptr;
  bool flag = true;
  cin >> n;
  cin >> str >> ptr;
  a = 0, b = 0;
  vector<long long> v;
  long long o = 0, z = 0;
  long long ans = 0;
  for (i = 0; i < n; i++) {
    a += (str[i] == '1');
    b += (ptr[i] == '1');
    if (str[i] == ptr[i]) continue;
    if (str[i] == '0' && ptr[i] == '1')
      z++, o--;
    else {
      o++;
      z--;
    }
    z = max(z, 0ll);
    o = max(o, 0ll);
    ans = max(ans, abs(z - o));
  }
  if (a == b) {
    cout << ans << endl;
  } else
    cout << -1 << endl;
}
