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 power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % 998244353;
    y = y / 2;
    if (y != 0) x = (x * x) % 998244353;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, c = 0, i, j, ct = 0, ans = 0, x, y;
  cin >> n;
  string s, t;
  cin >> s >> t;
  set<int> v, w;
  for (i = 0; i < n; i++) {
    if (s[i] == '0' && t[i] == '1') {
      v.insert(i);
    } else if (s[i] == '1' && t[i] == '0') {
      w.insert(i);
    }
    if (s[i] == '1') c++;
    if (t[i] == '1') ct++;
  }
  if (ct != c)
    cout << "-1";
  else {
    if (v.size() == 0) {
      cout << "0";
      return 0;
    }
    int flag = 1;
    while (v.size() || w.size()) {
      if (flag) {
        x = *v.begin();
        y = *w.begin();
        flag = 0;
        v.erase(v.begin());
        w.erase(w.begin());
      }
      if (x < y) {
        auto it = v.lower_bound(y);
        if (it == v.end()) {
          ans++;
          flag = 1;
        } else {
          x = *it;
          auto it1 = w.lower_bound(x);
          if (it1 == w.end()) {
            ans++;
            flag = 1;
          } else {
            v.erase(it);
            y = *it1;
            w.erase(it1);
          }
        }
      } else {
        auto it = w.lower_bound(x);
        if (it == w.end()) {
          ans++;
          flag = 1;
        } else {
          y = *it;
          auto it1 = v.lower_bound(y);
          if (it1 == v.end()) {
            ans++;
            flag = 1;
          } else {
            x = *it1;
            v.erase(it1);
            w.erase(it);
          }
        }
      }
    }
    if (flag == 0) ans++;
    cout << ans;
  }
  return 0;
}
