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;
long long n;
set<long long> zz;
set<long long> oo;
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  string s1, s2;
  cin >> s1 >> s2;
  long long tot1z = 0, tot11 = 0;
  long long tot2z = 0, tot21 = 0;
  for (int i = 0; i < s1.length(); i++) {
    if (s1[i] == '0') {
      tot1z++;
    } else {
      tot11++;
    }
    if (s2[i] == '0') {
      tot2z++;
    } else
      tot21++;
    if (s1[i] != s2[i]) {
      if (s1[i] == '0')
        zz.insert(i);
      else
        oo.insert(i);
    }
  }
  if (tot21 != tot11 || tot1z != tot2z) {
    cout << -1 << endl;
    return 0;
  }
  long long tot = 0;
  while (!oo.empty() && !zz.empty()) {
    tot++;
    long long sto = *oo.begin();
    long long stz = *zz.begin();
    if (sto < stz) {
      long long b = sto;
      while (!oo.empty() && !zz.empty()) {
        if (oo.lower_bound(b) != oo.end()) {
          long long d = *oo.lower_bound(b);
          if (zz.lower_bound(d) != zz.end()) {
            long long x = *zz.lower_bound(d);
            oo.erase(d);
            zz.erase(x);
            b = x;
          } else
            break;
        } else
          break;
      }
    } else {
      long long b = stz;
      while (!oo.empty() && !zz.empty()) {
        if (zz.lower_bound(b) != zz.end()) {
          long long d = *zz.lower_bound(b);
          if (oo.lower_bound(d) != oo.end()) {
            long long x = *oo.lower_bound(d);
            zz.erase(d);
            oo.erase(x);
            b = x;
          } else
            break;
        } else
          break;
      }
    }
  }
  cout << tot << endl;
  return 0;
}
