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 long long int MOD = 1e9 + 7;
inline long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
inline long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
inline long long int add(long long int a, long long int b) {
  return ((a % MOD) + (b % MOD)) % MOD;
}
inline long long int multi(long long int a, long long int b) {
  return ((a % MOD) * (b % MOD)) % MOD;
}
inline long long int sub(long long int a, long long int b) {
  a %= MOD;
  b %= MOD;
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
inline long long int power(long long int a, long long int b) {
  a %= MOD;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = multi(res, a);
    }
    a = multi(a, a);
    b >>= 1;
  }
  return res;
}
bool isPrime(long long int x) {
  if (x <= 4 || x % 2 == 0 || x % 3 == 0) {
    return x == 2 || x == 3;
  }
  for (long long int i = 5; i * i <= x; i += 6) {
    if (x % i == 0 || x % (i + 2) == 0) {
      return 0;
    }
  }
  return 1;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  string s, t;
  cin >> s >> t;
  long long int c1 = 0, c2 = 0;
  vector<long long int> res;
  long long int ans = 0;
  bool flag = 0;
  for (long long int i = 0; i < n; i++) {
    if (s[i] == '1') c1++;
    if (t[i] == '1') c2++;
  }
  if (c1 != c2) {
    cout << -1;
    return 0;
  }
  long long int pos = 0, neg = 0;
  for (long long int i = 0; i < n; i++) {
    if (s[i] == '1' && t[i] == '0') {
      if (neg > 0) {
        pos++;
        neg--;
      } else
        pos++;
    } else if (s[i] == '0' && t[i] == '1') {
      if (pos > 0) {
        neg++;
        pos--;
      } else
        neg++;
    }
  }
  cout << neg + pos;
}
