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;
template <typename T>
void out(T x) {
  cout << x << endl;
  exit(0);
}
using ll = long long;
const ll mod = 1e9 + 7;
const int maxn = 1e6 + 5;
void check(string s, string t) {
  sort(s.begin(), s.end());
  sort(t.begin(), t.end());
  if (s != t) out(-1);
}
int n;
int a[maxn], b[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s, t;
  cin >> n >> s >> t;
  check(s, t);
  for (int i = 0; i < n; i++) {
    a[i] = (s[i] == '1');
    b[i] = (t[i] == '1');
  }
  vector<int> v;
  for (int i = 0; i < n; i++) {
    if (a[i] == 0 && b[i] == 1) {
      v.push_back(0);
    } else if (a[i] == 1 && b[i] == 0) {
      v.push_back(1);
    }
  }
  vector<int> last(2, 0);
  for (int i : v) {
    if (last[i ^ 1] == 0) {
      last[i ^ 1]++;
    }
    last[i ^ 1]--;
    last[i]++;
  }
  cout << last[0] + last[1] << endl;
  return 0;
}
