#include<iostream>
#include<string>
#include<algorithm>    
#include<cmath>
#include<ctime>
#include<map>
#include<vector>
#include<math.h>
#include<stdio.h>
#include<stack>
#include<queue>
#include<tuple>
#include<cassert>
#include<set>
#include<bitset>
#include<functional>
#include <fstream>
//#include<bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#define rep(i, x) for(ll i = 0; i < x; i++)
#define rep2(i, x) for(ll i = 1; i <= x; i++)
#define rep3(i, x, y) for(ll i = x; i < y; i++)
#define rep4(i, x) for(ll i = x; i >= 0; i--)
#define all(a) (a).begin(),(a).end()
#define puts(x) cout << (x) << "\n"
using ll = long long;
using ld = long double;
using namespace std;
const ll INF = 1000000000000000000;
const int intINF = 1000000000;
const ll mod = 1000000007;
const ll MOD = 998244353;
const ld pi = 3.141592653589793238;
//const ld EPS = 1e-9;

bool isprime(int p) {
	if (p == 1) return false;
	for (int i = 2; i < p; i++) {
		if (p % i == 0) return false;
	}
	return true;
}
ll gcd(ll a, ll b) {
	if (a < b)swap(a, b);
	if (a % b == 0)return b;
	return gcd(b, a % b);
}
//ax + by = cが整数解をもつための必要十分条件は c が gcd(a,b) で割り切れること。
// 返り値: a と b の最大公約数
// ax + by = gcd(a, b) を満たす (x, y) が格納される
//main関数内に extGCD(a, b, x, y); でx, yに解が格納
ll extGCD(ll a, ll b, ll& x, ll& y) {
	if (b == 0) {
		x = 1;
		y = 0;
		return a;
	}
	ll d = extGCD(b, a % b, y, x);
	y -= a / b * x;
	return d;
}
ll lcm(ll a, ll b) {
	return a / gcd(a, b) * b;
}
ll keta(ll n) {
	ll res = 0;
	while (n >= 1) {
		res += n % 10; n /= 10;
	}
	return res;
}
ll modpow(ll x, ll y, ll m) {
	ll res = 1;
	while (y) {
		if (y % 2) { res *= x; res %= m; }
		x = x * x % m; y /= 2;
	}
	return res;
}
ll kaijyo[600005];
void nCkinit(ll n, ll m) {
	ll cnt = 1; kaijyo[0] = 1;
	for (int h = 1; h <= n; h++) { cnt *= h; cnt %= m; kaijyo[h] = cnt; }
}
ll nCk(ll n, ll k, ll m) {
	/*ll a = 1, b = 1;
	for (int h = 1; h <= n; h++) { a *= h; a %= m; }
	for (int h = 1; h <= k; h++) { b *= h; b %= m; }
	for (int h = 1; h <= n - k; h++) { b *= h; b %= m; }*/
	ll a = kaijyo[n], b = kaijyo[k] * kaijyo[n - k] % m;
	return a * modpow(b, m - 2, m) % m;
}
//printf("%.10f\n", n);
typedef pair <ll, ll> P;
typedef pair <ll, string> pp;
ll dx[4] = { 1, 0, -1, 0 }, dy[4] = { 0, 1, 0, -1 };
struct edge { ll to, cost; };
struct status {
	ll d;
	ll c;
	ll s;

	bool operator<(const status& rhs) const { return d < rhs.d; };
	bool operator>(const status& rhs) const { return d > rhs.d; };
};



int main() {
	ios::sync_with_stdio(false);
	std::cin.tie(nullptr);
	//cout << fixed << setprecision(15);

	//input
	ll n; string s, t; cin >> n >> s >> t;
	ll zero = 0;
	rep(i, n) { if (s[i] == '0') { zero++; } }
	rep(i, n) { if (t[i] == '0') { zero--; } }
	if (zero != 0) {
		cout << "-1" << endl; return 0;
	}
	ll sone = 0, tone = 0, l = -1, ans = 0;
	rep(i, n) {
		if (s[i] == '1') { sone++; }
		if (t[i] == '1') { tone++; }
		if (sone == tone) {
			if (sone != 0) ans += (i - l) - sone;
			sone = 0; tone = 0; l = i;
		}
	}
	cout << ans << endl;
	return 0;
}
#include <bits/stdc++.h>
using namespace::std;

int main() {
    int N;
    cin >> N;
    string s, t;
    cin >> s >> t;
    int s0 = 0;
    int t0 = 0;
    int ans = 0;
    for(int i = 0; i < N; i++) {
        if(s[i] == '0' && t[i] == '1') {
            ans += 1;
            s0 += 1;
        } else if(s[i] == '1' && t[i] == '0') {
            t0 += 1;
        } else if(s0 != t0 && s[i] == '0') {
            ans += 1;
        }
    }
    if(s0 != t0) {
        cout << -1 << endl;
        return 0;
    } 
    cout << ans << endl;
}