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;
int mod_pow(int num, int power) {
  int test;
  for (test = 1; power; power >>= 1) {
    if (power & 1) test = (test * num) % 1000000007;
    num = (num * num) % 1000000007;
  }
  return test;
}
void fastscan(int &number) {
  bool negative = false;
  register int c;
  number = 0;
  c = getchar();
  if (c == '-') {
    negative = true;
    c = getchar();
  }
  for (; (c > 47 && c < 58); c = getchar()) number = number * 10 + c - 48;
  if (negative) number *= -1;
}
bool cmp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  return (a.first.second < b.first.second);
}
vector<int> con[3000005];
int A[3000005], B[3000005];
int N, K;
void solve(int test_num) {
  cin >> N;
  string str;
  cin >> str;
  int su = 0;
  for (int i = 0; i < N; i++) {
    A[i] = (str[i] - '0');
    su += A[i];
  }
  cin >> str;
  for (int i = 0; i < N; i++) {
    B[i] = (str[i] - '0');
    su -= B[i];
  }
  if (su != 0) {
    cout << "-1" << endl;
    return;
  }
  vector<int> vc;
  for (int i = 0; i < N; i++) {
    if (A[i] == 0 && B[i] == 1) vc.push_back(-1);
    if (A[i] == 1 && B[i] == 0) vc.push_back(1);
  }
  if (vc.size() == 0) {
    cout << 0 << endl;
    return;
  }
  int ans = 0;
  int cu = 0;
  for (int i = 0; i < vc.size(); i++) {
    cu += vc[i];
    cu = max(cu, 0);
    ans = max(ans, cu);
  }
  cu = 0;
  for (int i = 0; i < vc.size(); i++) {
    cu += -1 * vc[i];
    cu = max(cu, 0);
    ans = max(ans, cu);
  }
  cout << ans << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int T;
  T = 1;
  for (int i = 0; i < T; i++) {
    solve(i + 1);
  }
  return (0);
}
