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;
vector<int> start, targ;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  string s, t;
  cin >> s >> t;
  int s_0 = 0, s_1 = 0, t_0 = 0, t_1 = 0;
  for (auto &x : s) {
    if (x == '0') {
      start.push_back(0);
      s_0++;
    } else {
      start.push_back(1);
      s_1++;
    }
  }
  for (auto &x : t) {
    if (x == '0') {
      targ.push_back(0);
      t_0++;
    } else {
      targ.push_back(1);
      t_1++;
    }
  }
  if (s_0 != t_0 || s_1 != t_1) {
    cout << -1 << '\n';
    return 0;
  }
  set<int> zero_one, one_zero;
  for (int i = 0; i < n; i++) {
    if (s[i] == '0' && t[i] == '1') zero_one.insert(i);
    if (s[i] == '1' && t[i] == '0') one_zero.insert(i);
  }
  int ans = 0;
  while (!zero_one.empty()) {
    ans++;
    int beg1 = *zero_one.begin(), beg2 = *one_zero.begin();
    bool is_begin_1_0, cur_1_0;
    int cur_ind = min(beg1, beg2);
    if (beg1 < beg2) {
      is_begin_1_0 = false;
      cur_1_0 = false;
    } else {
      is_begin_1_0 = true;
      cur_1_0 = true;
    }
    vector<int> changes{cur_ind};
    while (true) {
      if (cur_1_0) {
        auto it = zero_one.upper_bound(cur_ind);
        if (it == zero_one.end())
          break;
        else {
          cur_ind = *it;
          changes.push_back(cur_ind);
          cur_1_0 = !cur_1_0;
        }
      } else {
        auto it = one_zero.upper_bound(cur_ind);
        if (it == one_zero.end())
          break;
        else {
          cur_ind = *it;
          changes.push_back(cur_ind);
          cur_1_0 = !cur_1_0;
        }
      }
    }
    if (is_begin_1_0 == cur_1_0) {
      changes.pop_back();
    }
    for (auto &x : changes) {
      zero_one.erase(x);
      one_zero.erase(x);
    }
  }
  cout << ans << '\n';
  return 0;
}
