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;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using ld = long double;
const int mod = 1e9 + 7;
const int inf = 1e9 + 11;
const ll INF = 2e18;
const int N = 2e6 + 66;
const int M = 5005;
string s, t;
int a[N], b[N];
set<int> pos[2];
void solve() {
  int n;
  cin >> n >> s >> t;
  for (int i = 1; i <= n; ++i) {
    a[i] = (s[i - 1] == '1');
    b[i] = (t[i - 1] == '1');
    if (a[i] != b[i]) {
      pos[a[i]].insert(i);
    }
  }
  sort(s.begin(), s.end());
  sort(t.begin(), t.end());
  if (s != t) {
    cout << -1;
    return;
  }
  int cnt = 0;
  while (pos[0].size() && pos[1].size()) {
    int now = *pos[0].begin(), type = 0;
    if (now > *pos[1].begin()) {
      now = *pos[1].begin();
      type = 1;
    }
    pos[type].erase(now);
    type ^= 1;
    int cur = 1;
    while (pos[type].size() && *pos[type].rbegin() >= now) {
      auto p = pos[type].lower_bound(now);
      now = *p;
      pos[type].erase(p);
      type ^= 1;
      cur ^= 1;
    }
    if (cur) {
      type ^= 1;
      pos[type].insert(now);
    }
    ++cnt;
  }
  if (pos[0].size() || pos[1].size()) {
    cout << -1;
    return;
  }
  cout << cnt;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  int tt = 1;
  while (tt--) {
    solve();
    cout << "\n";
  }
}
