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 int INF = 2147483647;
const long long LONGINF = 9223372036854775807;
const long long MOD = 998244353;
string s, t;
set<int> change0, change1;
void solve() {
  int n;
  cin >> n >> s >> t;
  for (int i = 0; i < n; ++i) {
    if (s[i] != t[i] && s[i] == '0')
      change0.insert(i);
    else if (s[i] != t[i])
      change1.insert(i);
  }
  if (change0.size() != change1.size()) {
    cout << -1 << '\n';
    return;
  }
  int res = 0;
  while (!change0.empty()) {
    ++res;
    if (*change0.begin() > *change1.begin()) change0.swap(change1);
    int current = *change0.begin();
    while (true) {
      auto it0 = change0.lower_bound(current);
      if (it0 == change0.end()) break;
      auto it1 = change1.lower_bound(*it0);
      if (it1 == change1.end()) break;
      change0.erase(it0);
      current = *it1;
      change1.erase(it1);
    }
  }
  cout << res << '\n';
}
signed main() {
  cin.tie(0);
  cout.tie(0);
  std::ios_base::sync_with_stdio(0);
  bool mult = false;
  if (mult) {
    int t;
    cin >> t;
    while (t--) solve();
  } else
    solve();
}
