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>
#pragma GCC optimize("O3")
using namespace std;
long long n, i, j, cnt, ans, m;
string second, t;
deque<pair<long long, long long>> v;
signed main() {
  iostream::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> second >> t;
  for (char h : second) cnt += h - '0';
  for (char h : t) cnt -= h - '0';
  if (cnt != 0) {
    cout << "-1";
    exit(0);
  }
  for (i = 0; i < n; i++) {
    if (second[i] == t[i]) continue;
    j = i + 1;
    cnt = 1;
    while (j < n) {
      if (second[j] == t[j]) {
        j++;
        continue;
      }
      if (second[i] != second[j]) break;
      cnt++;
      j++;
    }
    v.push_back({second[i] - '0', cnt});
    i = j - 1;
  }
  while (v.size() > 0) {
    if (v[0].first == 1) {
      if (v.back().first == 1) {
        v[0].second += v.back().second;
        v.pop_back();
      }
    } else {
      if (v.back().first == 0)
        v.back().second += v[0].second;
      else
        v.push_back(v[0]);
      v.pop_front();
    }
    m = 1e9;
    for (auto x : v) m = min(m, x.second);
    ans += m;
    for (i = 0; i < v.size(); i++) v[i].second -= m;
    deque<pair<long long, long long>> v1 = v;
    v.clear();
    for (i = 0; i < v1.size(); i++) {
      if (v1[i].second == 0) continue;
      j = i + 1;
      cnt = v1[i].second;
      while (j < v1.size()) {
        if (v1[j].second == 0) {
          j++;
          continue;
        }
        if (v1[i].first != v1[j].first) break;
        cnt += v1[j].second;
        j++;
      }
      v.push_back({v1[i].first, cnt});
      i = j - 1;
    }
  }
  cout << ans;
  return 0;
}
