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;
void setIO(const string &name = "") {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  if (name.length()) {
    freopen((name + ".in").c_str(), "r", stdin);
    freopen((name + ".out").c_str(), "w", stdout);
  }
}
template <typename T>
void read(vector<T> &a) {
  for (auto &x : a) cin >> x;
}
template <typename T>
void read(vector<T> &a, long long n) {
  a.resize(n);
  for (auto &x : a) cin >> x;
}
template <typename T>
void print(vector<T> &a) {
  for (const auto &x : a) cout << x << ' ';
  cout << '\n';
}
void MOD(long long &x) {
  x %= 1000000007;
  if (x < 0) x += 1000000007;
}
long long gcd(long long a, long long b) { return !a ? b : gcd(b % a, a); }
long long n;
string s, t;
vector<char> disp;
int32_t main() {
  setIO();
  cin >> n;
  cin >> s >> t;
  long long nums0 = 0, nums1 = 0;
  long long numt0 = 0, numt1 = 0;
  for (auto(i) = 0; i < n; i++) {
    if (s[i] != t[i]) {
      disp.emplace_back(s[i]);
      if (s[i] == '1') {
        nums1++;
        numt0++;
      } else {
        nums0++;
        numt1++;
      }
    }
  }
  if (nums0 != numt0 or nums1 != numt1) {
    cout << "-1\n";
    return 0;
  }
  if (disp.empty()) {
    cout << "0\n";
    return 0;
  }
  long long ans = 0;
  long long end0 = 0, end1 = 0;
  for (auto x : disp) {
    if (x == '0') {
      if (end1 == 0) {
        ans++;
      } else {
        end1--;
      }
      end0++;
    } else {
      if (end0 == 0) {
        ans++;
      } else {
        end0--;
      }
      end1++;
    }
  }
  cout << ans << '\n';
  return 0;
}
