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;
priority_queue<long long int, vector<long long int>, greater<long long int> >
    ti;
vector<long long int> p[300005], a(300005, 0), b(200005, 0), f(400005, 0),
    d(400005, 0), q[200005];
long long int vis[200005];
map<pair<long long int, long long int>, long long int> mp;
set<pair<long long int, long long int> > s;
set<long long int> st;
map<long long int, long long int> np, v;
queue<long long int> qy;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int bpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res % 1000000007;
}
long long int fact(long long int i) {
  f[0] = 1;
  for (long long int k = 1; k <= i; k++) {
    (f[k] = f[k - 1] * k) %= 1000000007;
  }
  return f[i];
}
long long int isprime(long long int n) {
  if (n == 1) return 0;
  for (long long int i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return 0;
  return 1;
}
bool cmp(long long int i, long long int j) { return p[i].size() < p[j].size(); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  string s, t;
  cin >> s >> t;
  long long int c1 = 0, c2 = 0;
  for (long long int i = 0; i < n; i++) {
    if (s[i] == '1') c1++;
  }
  for (long long int i = 0; i < n; i++) {
    if (t[i] == '1') c2++;
  }
  if (c1 != c2) {
    cout << -1 << "\n";
    exit(0);
  }
  map<long long int, long long int> mp;
  for (long long int i = 0; i < n; i++) {
    if (s[i] != t[i] && s[i] == '1') mp[i] = 1;
    if (s[i] != t[i] && s[i] == '0') mp[i] = -1;
  }
  c1 = 0;
  c2 = 0;
  char c = '.';
  for (long long int i = 0; i < n; i++) {
    c1 += mp[i];
    if (c1 < 0) c1 = 0;
    c2 = max(c2, c1);
  }
  c1 = 0;
  for (long long int i = 0; i < n; i++) {
    c1 += (-mp[i]);
    if (c1 < 0) c1 = 0;
    c2 = max(c2, c1);
  }
  cout << c2 << "\n";
}
