You are given strings s and t, both of length N. s and t consist of `0` and `1`. Additionally, in these strings, the same character never occurs three or more times in a row.

You can modify s by repeatedly performing the following operation:

* Choose an index i (1 \leq i \leq N) freely and invert the i-th character in s (that is, replace `0` with `1`, and `1` with `0`), under the condition that the same character would not occur three or more times in a row in s after the operation.



Your objective is to make s equal to t. Find the minimum number of operations required.

Constraints

* 1 \leq N \leq 5000
* The lengths of s and t are both N.
* s and t consists of `0` and `1`.
* In s and t, the same character never occurs three or more times in a row.

Input

Input is given from Standard Input in the following format:


N
s
t


Output

Find the minimum number of operations required to make s equal to t. It can be proved that the objective is always achievable in a finite number of operations.

Examples

Input

4
0011
0101


Output

4


Input

1
0
0


Output

0


Input

8
00110011
10101010


Output

10
// #pragma GCC target("avx")  // CPU 処理並列化
// #pragma GCC optimize("O3")  // CPU 処理並列化
// #pragma GCC optimize("unroll-loops")  // 条件処理の呼び出しを減らす
// #define BEGIN_STACK_EXTEND(size) void * stack_extend_memory_ = malloc(size);void * stack_extend_origin_memory_;char * stack_extend_dummy_memory_ = (char*)alloca((1+(int)(((long long)stack_extend_memory_)&127))*16);*stack_extend_dummy_memory_ = 0;asm volatile("mov %%rsp, %%rbx\nmov %%rax, %%rsp":"=b"(stack_extend_origin_memory_):"a"((char*)stack_extend_memory_+(size)-1024));
// #define END_STACK_EXTEND asm volatile("mov %%rax, %%rsp"::"a"(stack_extend_origin_memory_));free(stack_extend_memory_);

#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<stack>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#include<stdlib.h>
#include<cassert>
#include<time.h>
#include<bitset>
#include<numeric>
#include<unordered_set>
#include<unordered_map>
#include<complex>
using namespace std;
const long long mod=1000000007;
const long long inf=mod*mod;
const long long d2=(mod+1)/2;
const long double EPS=1e-9;
const long double INF=1e+10;
const long double PI=acos(-1.0);
const int C_SIZE = 3100000;
const int UF_SIZE = 3100000;
namespace{
	long long fact[C_SIZE];
	long long finv[C_SIZE];
	long long inv[C_SIZE];
	long long Comb(int a,int b){
	 	if(a<b||b<0)return 0;
	 	return fact[a]*finv[b]%mod*finv[a-b]%mod;
	}
	void init_C(int n){
		fact[0]=finv[0]=inv[1]=1;
		for(int i=2;i<n;i++){
			inv[i]=(mod-(mod/i)*inv[mod%i]%mod)%mod;
		}
		for(int i=1;i<n;i++){
			fact[i]=fact[i-1]*i%mod;
			finv[i]=finv[i-1]*inv[i]%mod;
		}
	}
	long long pw(long long a,long long b){
		if(a<0LL)return 0;
		if(b<0LL)return 0;
		long long ret=1;
		while(b){
			if(b%2)ret=ret*a%mod;
			a=a*a%mod;
			b/=2;
		}
		return ret;
	}
	long long pw_mod(long long a,long long b,long long M){
		if(a<0LL)return 0;
		if(b<0LL)return 0;
		long long ret=1;
		while(b){
			if(b%2)ret=ret*a%M;
			a=a*a%M;
			b/=2;
		}
		return ret;
	}
	int pw_mod_int(int a,int b,int M){
		if(a<0)return 0;
		if(b<0)return 0;
		int ret=1;
		while(b){
			if(b%2)ret=(long long)ret*a%M;
			a=(long long)a*a%M;
			b/=2;
		}
		return ret;
	}
	int ABS(int a){return max(a,-a);}
	long long ABS(long long a){return max(a,-a);}
	double ABS(double a){return max(a,-a);}
	int sig(double r) { return (r < -EPS) ? -1 : (r > +EPS) ? +1 : 0; }
	int UF[UF_SIZE];
	void init_UF(int n){
		for(int i=0;i<n;i++)UF[i]=-1;
	}
	int FIND(int a){
		if(UF[a]<0)return a;
		return UF[a]=FIND(UF[a]);
	}
	void UNION(int a,int b){
		a=FIND(a);b=FIND(b);if(a==b)return;
		// if(UF[a]>UF[b])swap(a,b);
		UF[a]+=UF[b];UF[b]=a;
	}
}
// ここから編集しろ

char S[5100];
char T[5100];
vector<int>p;
vector<int>q;
int dp[5100][5100][2];
int main(){
	int a;scanf("%d",&a);
	scanf("%s%s",S,T);
	for(int i=0;i+1<a;i++){
		if(S[i]==S[i+1]){
			p.push_back(i+1);
		}
		if(T[i]==T[i+1]){
			q.push_back(i+1);
		}
	}
	for(int i=0;i<=p.size();i++){
		for(int j=0;j<=q.size();j++)for(int k=0;k<2;k++)
			dp[i][j][k]=mod;
	}
	dp[0][0][0]=0;
	for(int i=0;i<=p.size();i++){
		for(int j=0;j<=q.size();j++){
			for(int k=0;k<2;k++){
				if(dp[i][j][k]>=mod)continue;
				if(i<p.size()){
					if(j==0){
						dp[i+1][j][!k]=min(dp[i+1][j][!k],dp[i][j][k]+p[i]);
					}
					if(j<q.size()){
						dp[i+1][j+1][k]=min(dp[i+1][j+1][k],dp[i][j][k]+ABS(p[i]-q[j]));
					}else if(j==q.size()){
						dp[i+1][j][k]=min(dp[i+1][j][k],dp[i][j][k]+a-p[i]);
					}
				}
				if(j<q.size()){
					if(i==0){
						dp[i][j+1][!k]=min(dp[i][j+1][!k],dp[i][j][k]+q[j]);
					}
					if(i<p.size()){
					}else if(i==p.size()){
						dp[i][j+1][k]=min(dp[i][j+1][k],dp[i][j][k]+a-q[j]);
					}
				}

			}
		}
	}
	int ret=mod;
	int t=(S[0]!=T[0]);
	ret=min(dp[p.size()][q.size()][t],dp[p.size()][q.size()][!t]+a);
	printf("%d\n",ret);
}