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 mod = 1e9 + 7;
long long power(long long a, long long b) {
  if (b == 0) return 1;
  long long res = power(a, b / 2);
  long long ans = res * res;
  if (b % 2 == 0) return ans;
  return a * ans;
}
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  string s, t;
  cin >> s >> t;
  const int N = 1e6 + 5;
  bitset<N> b1(s), b2(t), b3;
  if (b1.count() != b2.count()) {
    cout << -1 << "\n";
    return 0;
  }
  int ones = b1.count();
  int ans = min(ones, n - ones);
  b3 = b1 & b2;
  int common_ones = b3.count();
  ans = min(ans, ones - common_ones);
  string x = "";
  for (int i = 0; i < n; i++) {
    if (s[i] != t[i]) {
      x += s[i];
    }
  }
  int len = x.size();
  if (!len) {
    cout << 0 << "\n";
    return 0;
  }
  int cnt = 0;
  int z = 0, o = 0;
  for (int i = 0; i < len; i++) {
    if (x[i] == '0') {
      if (o) o--;
      z++;
    } else {
      if (z) z--;
      o++;
    }
  }
  cnt = o + z;
  ans = min(ans, cnt);
  cout << ans << "\n";
  return 0;
}
