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.
import io
import os

from collections import Counter, defaultdict, deque


def solve(N, S, T):
    S0 = S.count("0")
    T0 = T.count("0")
    S1 = S.count("1")
    T1 = T.count("1")
    if S0 != T0 or S1 != T1:
        return -1

    consect = []
    for s, t in zip(S, T):
        if s != t:
            if not consect:
                consect.append([s, 1])
            else:
                if consect[-1][0] == s:
                    consect[-1][1] += 1
                else:
                    consect.append([s, 1])

    steps = 0
    while consect:
        nextConsect = []
        for ch, count in consect:
            if count > 1:
                if not nextConsect:
                    nextConsect.append([ch, count - 1])
                else:
                    if nextConsect[-1][0] == ch:
                        nextConsect[-1][1] += count - 1
                    else:
                        nextConsect.append([ch, count - 1])
        consect = nextConsect

        steps += 1

    return steps

'''
for i in range(2 ** 5):
    for j in range(2 ** 5):
        solve(5, bin(i)[2:].rjust(5, '0'), bin(j)[2:].rjust(5, '0'))

'''

if __name__ == "__main__":
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

    (N,) = [int(x) for x in input().split()]
    S = input().decode().rstrip()
    T = input().decode().rstrip()
    ans = solve(N, S, T)
    print(ans)
