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 q, n, c, x, r, a1[2], a2[2], a[1000000][2], c0, c1;
string s1, s2;
bool b[1000001];
long long dfs(long long xf) {
  if (!b[xf]) return xf;
  return a[xf][s1[xf] - '0'] = dfs(a[xf][s1[xf] - '0']);
}
int main() {
  cin >> n;
  cin >> s1 >> s2;
  for (int i = 0; i < n; i++) {
    a1[s1[i] - '0']++;
    a2[s2[i] - '0']++;
  }
  if (a1[0] != a2[0]) {
    cout << -1;
    return 0;
  }
  c0 = n;
  c1 = n;
  for (int i = n - 1; i >= 0; i--) {
    a[i][0] = c0;
    a[i][1] = c1;
    if (s1[i] != s2[i]) {
      if (s1[i] - '0' == 0)
        c0 = i;
      else
        c1 = i;
    }
  }
  for (int i = 0; i < n; i++) {
    if (s1[i] != s2[i] && !b[i]) {
      c = (s1[i] - '0' + 1) % 2;
      x = i;
      while (x != n) {
        b[x] = 1;
        if (b[a[x][c]]) a[x][c] = dfs(a[x][c]);
        x = a[x][c];
        c = (c + 1) % 2;
      }
      r++;
    }
  }
  cout << r;
  return 0;
}
