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;
namespace io {
inline char nc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
template <class I>
inline void fr(I &num) {
  num = 0;
  register char c = nc();
  register int p = 1;
  while (c < '0' || c > '9') c == '-' ? p = -1, c = nc() : c = nc();
  while (c >= '0' && c <= '9') num = num * 10 + c - '0', c = nc();
  num *= p;
}
template <class I>
inline I Max(I p, I q) {
  return p > q ? p : q;
}
template <class I>
inline I Min(I p, I q) {
  return p < q ? p : q;
}
template <class I>
inline I A(I x) {
  return x < 0 ? -x : x;
}
template <class I>
inline void sp(I &p, I &q) {
  I x = p;
  p = q, q = x;
}
template <class I>
inline void ckMax(I &p, I q) {
  p = (p > q ? p : q);
}
template <class I>
inline void ckMin(I &p, I q) {
  p = (p < q ? p : q);
}
};  // namespace io
using io ::A;
using io ::ckMax;
using io ::ckMin;
using io ::fr;
using io ::Max;
using io ::Min;
using io ::nc;
using io ::sp;
const int N = 1000005;
int T, n, ans1, ans2;
char s[N], t[N];
int main() {
  scanf("%d%s%s", &n, s + 1, t + 1);
  register int i, now = 0, cn0 = 0, cn1 = 0;
  for (i = 1; i <= n; ++i)
    cn0 += (s[i] == '0'), cn0 -= (t[i] == '0'), cn1 += (s[i] == '1'),
        cn1 -= (t[i] == '1');
  if (cn0 || cn1) return puts("-1"), 0;
  for (i = 1; i <= n; ++i) {
    if (s[i] != t[i]) now += (s[i] == '0' ? 1 : -1);
    ans1 = Max(ans1, now), ans2 = Min(ans2, now);
  }
  printf("%d", ans1 - ans2);
  return 0;
}
