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;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  string s;
  cin >> s;
  string t;
  cin >> t;
  set<int> zero1;
  set<int> one0;
  for (int i = 0; i < n; i++) {
    if (s[i] == '1' && t[i] == '0') {
      one0.emplace(i);
    } else if (s[i] == '0' && t[i] == '1') {
      zero1.emplace(i);
    }
  }
  if (one0.size() != zero1.size()) {
    cout << -1 << endl;
    return 0;
  }
  int ans = 0;
  while (zero1.size() > 0) {
    bool f = 1;
    vector<int> rem1, rem0;
    int curr = -1;
    if (one0.size() > 0 && zero1.size() > 0) {
      int ind1 = *one0.begin();
      int ind2 = *zero1.begin();
      f = ind1 < ind2;
    }
    while (one0.size() > 0 && zero1.size() > 0) {
      if (f) {
        auto ind = one0.lower_bound(curr);
        if (ind == one0.end()) {
          break;
        }
        curr = *ind;
        rem1.push_back(curr);
        f = !f;
      } else {
        auto ind = zero1.lower_bound(curr);
        if (ind == zero1.end()) {
          break;
        }
        curr = *ind;
        rem0.push_back(curr);
        f = !f;
      }
    }
    int minn = min(rem0.size(), rem1.size());
    if (minn == 0) break;
    ans += 1;
    for (int i = 0; i < minn; i++) {
      zero1.erase(rem0[i]);
      one0.erase(rem1[i]);
    }
  }
  ans += zero1.size();
  cout << ans << endl;
  return 0;
}
