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;
struct lex_compare {
 public:
  bool operator()(const pair<long long, long long>& lhs,
                  const pair<long long, long long>& rhs) const {
    if (lhs.second == rhs.second) return lhs.first < rhs.first;
    return (lhs.second > rhs.second);
  }
};
class trp {
 public:
  long long a, b, c;
  trp() {
    this->a = 0;
    this->b = 0;
    this->c = 0;
  }
  trp(long long a, long long b, long long c) {
    this->a = a;
    this->b = b;
    this->c = c;
  }
};
class node {
 public:
  long long index, a, l;
  long long d, h;
  node* mr_img;
  node* lf = NULL;
  node* rt = NULL;
  node* prt = NULL;
  vector<node*> v;
  node() {
    index = 0, a = 0, l = 0;
    d = 1, h = 0;
    mr_img = NULL;
    lf = NULL;
    rt = NULL;
    prt = NULL;
  }
  void adn(node* t) {
    v.push_back(t);
    this->l++;
  }
};
node* nnd(long long a) {
  node* t = new node();
  t->a = a;
  return t;
}
long long pw(long long a, long long b, long long m) {
  if (b == 0) return 1;
  if (b % 2 == 0) return (pw(a, b / 2, m) * pw(a, b / 2, m)) % m;
  return (a * ((pw(a, b / 2, m) * pw(a, b / 2, m)) % m)) % m;
}
long long gc(long long a, long long b) {
  if (b == 0) return a;
  return gc(b, a % b);
}
vector<string> splt(string x, char w) {
  vector<string> v;
  string s;
  int l = x.length();
  for (int i = 0; i < l; i++) {
    if (x[i] == w) {
      v.push_back(s);
      s = "";
    } else {
      s.push_back(x[i]);
    }
  }
  v.push_back(s);
  return v;
}
void ioe() {
  freopen("in_3.txt", "r", stdin);
  freopen("out_3.txt", "w", stdout);
  freopen("er_3.txt", "w", stderr);
}
void solve() {
  long long n, c = 0, l = 0, l1 = 0, l2 = 0;
  string s1, s2;
  cin >> n >> s1 >> s2;
  for (int i = 0; i < n; i++) {
    if (s1[i] == '1') c++;
    if (s2[i] == '1') c--;
  }
  if (c != 0) {
    cout << "-1\n";
    return;
  }
  vector<long long> s;
  for (int i = 0; i < n; i++) {
    if (s1[i] == '1' && s2[i] == '0')
      s.push_back(1), l++;
    else if (s2[i] == '1' && s1[i] == '0')
      s.push_back(0), l++;
  }
  if (l == 0) {
    cout << "0\n";
    return;
  }
  long long m1 = 1, c1 = 0, c2 = 0;
  for (int i = 0; i < 2 * l; i++) {
    if (s[i % l] == 0)
      c2++, c1--;
    else
      c2--, c1++;
    ;
    c1 = max(0ll, c1);
    c2 = max(0ll, c2);
    m1 = max(m1, max(c2, c1));
  }
  m1 = max(m1, max(c1, c2));
  cout << m1 << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t = 1;
  long long f = t;
  while (f--) {
    solve();
  }
}
