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.
n = int(input())
a = list(map(int,input()))
b = list(map(int,input()))
c = []
if sum(a) != sum(b):
    print(-1)
    exit()

for i in range(n):
    if a[i] != b[i]:
        c.append(a[i])

open1 = 0
open0 = 0
flag1 = False
flag0 = False
ans = 0
oooopen1 = 0
oooopen0 = 0
for i in c:
    if i == 0:
        if open1 > 0:
            open1 -= 1
        else:
            open0 += 1
            flag0 = True
    else:
        if open0 > 0:
            open0 -= 1
        else:
            open1 += 1
            flag1 = True
    #open0_ans = open0
    #open1_ans = open1
    oooopen1 = max(oooopen1, open1)
    oooopen0 = max(oooopen0, open0)

    ans = max(ans, oooopen1+oooopen0)
print(ans)


'''
104
00010011000000001001110111010001101011101010100000110000000010001101000101001011100110111101101000110000
00100010100111001010010000100110001001010100110010010101001011001000000001101001011101000101101100001111


[ 1, 1,  1,  0, 0, 0]
'''