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.
######################################################
############Created by Devesh Kumar###################
#############devesh1102@gmail.com####################
##########For CodeForces(Devesh1102)#################
#####################2020#############################
######################################################
import sys
input = sys.stdin.readline

# import sys
import heapq 
import copy
import math
import decimal
# import sys.stdout.flush as flush
# from decimal import *
#heapq.heapify(li) 
#
#heapq.heappush(li,4) 
#
#heapq.heappop(li)
#
# &	Bitwise AND Operator	10 & 7 = 2
# |	Bitwise OR Operator	10 | 7 = 15
# ^	Bitwise XOR Operator	10 ^ 7 = 13
 
# <<	Bitwise Left Shift operator	10<<2 = 40
# >>	Bitwise Right Shift Operator
# '''############ ---- Input Functions ---- #######Start#####'''
 

def inp():
	return(int(input()))
def inlt(): 
	return(list(map(int,input().split())))
def insr():
	s = input()
	return(list(s[:len(s) - 1]))
def insr2():
	s = input()
	return((s[:len(s) - 1]))
def invr():
	return(map(int,input().split()))
 ############ ---- Input Functions ---- #######End
 # #####   

def pr_list(a):
	print(*a, sep=" ")
def main():
	# tests =  inp()
	tests = 1
	mod = 1000000007
	limit = 10**18
	ans = 0
	start = {}
	for test in range(tests):
		n = inp()
		s = insr()
		t = insr()
		n_0t = 0
		n_1t = 0
		n_0s = 0
		n_1s = 0
		for i in range(n):
			if int(s[i] ) == 1:
				n_1s +=1
			else:
				n_0s +=1
			if int(t[i]) == 1:
				n_1t +=1
			else:
				n_0t += 1
		if n_0s!=n_0t  or n_1s!=n_1t:
			print(-1)
			continue
		one = 0
		zero = 0
		ans = 0
		for i in range(n):
			
			if s[i]!=t[i]:
				if s[i] == "0":
					one = one + 1
					zero = max(zero -1,0)
				elif s[i] == "1":
					one = max(one -1,0)
					zero = zero +1
			ans = max(ans,one,zero)
			# print(i,s[i],one,zero)
		print(ans)






		
		


if __name__== "__main__":
	main()