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;
int dx[] = {+1, -1, 0, 0};
int dy[] = {0, 0, +1, -1};
int n;
char s[1000005];
char t[1000005];
set<int> st[2];
set<int>::iterator it;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  scanf("%d", &n);
  scanf("%s", s);
  scanf("%s", t);
  int cnts0 = 0, cnts1 = 0, cntt0 = 0, cntt1 = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] != t[i]) {
      if (s[i] == '0')
        st[0].insert(i);
      else
        st[1].insert(i);
    }
    if (s[i] == '0')
      cnts0++;
    else
      cnts1++;
    if (t[i] == '0')
      cntt0++;
    else
      cntt1++;
  }
  if (cnts0 != cntt0) {
    cout << "-1" << endl;
    return 0;
  }
  st[0].insert(1000005);
  st[1].insert(1000005);
  int ans = 0;
  while (true) {
    int n0 = 1000005, n1 = 1000005;
    it = st[0].begin();
    n0 = *it;
    it = st[1].begin();
    n1 = *it;
    if (n0 == 1000005 && n1 == 1000005) break;
    int pos, cur = 0;
    if (n0 > n1)
      cur = 1, pos = n1;
    else
      cur = 0, pos = n0;
    ++ans;
    while (true) {
      st[cur].erase(pos);
      it = st[cur ^ 1].lower_bound(pos);
      if (*it == 1000005) break;
      pos = *it;
      cur ^= 1;
    }
  }
  printf("%d\n", ans);
  return 0;
}
