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 long long LINF = (long long)1e18 + 47;
const int INF = 2 * 1e9 + 47;
const int MOD = 1e9 + 7;
const int modulo = 1e8;
const int nax = (int)1e6 + 10;
const double EPS = 1e-7;
int c[nax];
int n;
vector<int> vec;
int get(int x) {
  int mx = 0;
  int cur = 0;
  for (int i = (0); i < (vec.size()); i++) {
    if (x == vec[i])
      cur++;
    else
      cur--;
    mx = max(mx, cur);
    if (cur < 0) cur = 0;
  }
  return mx;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  string a, b;
  cin >> a >> b;
  for (int i = (0); i < (n); i++) {
    if (a[i] != b[i]) {
      if (a[i] == '1')
        c[i] = 1;
      else
        c[i] = -1;
    }
  }
  int cnt = 0;
  for (int i = (0); i < (n); i++) {
    if (c[i] == 1)
      cnt++;
    else if (c[i] == -1)
      cnt--;
  }
  if (cnt) {
    cout << -1 << '\n';
    return 0;
  }
  for (int i = (0); i < (n); i++) {
    if (c[i] != 0) {
      vec.push_back(c[i]);
    }
  }
  int ans = 0;
  for (int i = (0); i < (vec.size()); i++) {
    int j = i;
    while (j < vec.size() && vec[j] == vec[i]) {
      j++;
    }
    int len = j - i;
    ans = max(ans, len);
    i = j - 1;
  }
  cout << max(get(-1), get(1)) << '\n';
}
