D. Two Strings Swaps

                                                                          D. Two Strings Swaps

                                                                             time limit per test

                                                                                2 seconds

                                                                      memory limit per test

                                                                            256 megabytes

                                                                                    input

                                                                           standard input

                                                                                 output

                                                                         standard output

You are given two strings aa and bb consisting of lowercase English letters, both of length nn. The characters of both strings have indices from 11 to nn, inclusive.

You are allowed to do the following changes:

• Choose any index ii (1≤i≤n1≤i≤n) and swap characters aiai and bibi;
• Choose any index ii (1≤i≤n1≤i≤n) and swap characters aiai and an−i+1an−i+1;
• Choose any index ii (1≤i≤n1≤i≤n) and swap characters bibi and bn−i+1bn−i+1.

Note that if nn is odd, you are formally allowed to swap a⌈n2⌉a⌈n2⌉ with a⌈n2⌉a⌈n2⌉ (and the same with the string bb) but this move is useless. Also you can swap two equal characters but this operation is useless as well.

You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.

In one preprocess move you can replace a character in aa with another character. In other words, in a single preprocess move you can choose any index ii (1≤i≤n1≤i≤n), any character cc and set ai:=cai:=c.

Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings aa and bb equal by applying some number of changes described in the list above.

Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess movesto the string bb or make any preprocess moves after the first change is made.

Input

The first line of the input contains one integer nn (1≤n≤1051≤n≤105) — the length of strings aa and bb.

The second line contains the string aa consisting of exactly nn lowercase English letters.

The third line contains the string bb consisting of exactly nn lowercase English letters.

Output

Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string aa equal to string bb with a sequence of changes from the list above.

Examples

input

Copy

7
abacaba
bacabaa

output

Copy

4

input

Copy

5
zcabd
dbacz

output

Copy

0

Note

In the first example preprocess moves are as follows: a1:=a1:='b', a3:=a3:='c', a4:=a4:='a' and a5:=a5:='b'. Afterwards, a=a="bbcabba". Then we can obtain equal strings by the following sequence of changes: swap(a2,b2)swap(a2,b2) and swap(a2,a6)swap(a2,a6). There is no way to use fewer than 44preprocess moves before a sequence of changes to make string equal, so the answer in this example is 44.

In the second example no preprocess moves are required. We can use the following sequence of changes to make aa and bb equal: swap(b1,b5)swap(b1,b5), swap(a2,a4)swap(a2,a4).

 

比较a[i],a[n-i-1],b[i],b[n-i-1]的各种情况，代码如下：

#include <cstdio>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;
#define rig n-i-1
const int MAXN = 1e5+20;
char a[MAXN], b[MAXN];
int vis[30];

int main() {

    int n, ans = 0;
    cin>>n;
    scanf("%s%s", a, b);
    for (int i = 0; i < n / 2; ++i) {
        map<char, int> s;
        s[a[i]]++; s[a[rig]]++;
        s[b[i]]++; s[b[rig]]++;//记录a[i],a[n-1-i],b[i],b[n-1-i]有几个相同的字符;
        if ((int)(s.size()) == 4)
            ans += 2;            //如果四个都不同，则必然要变两个,ans加2;
        else if ((int)(s.size()) == 3)
            ans += 1 + (a[i] == a[rig]); //如果有三个不同并且a[i]和a[n-1-i]相同,ans加2,fo否则加1;
        else if ((int)(s.size()) == 2)
            ans += s[a[i]] != 2;   //如果有两个不同并且a[i]的数量不是两个,则加1;
    }

    if ((n & 1) && a[n/2] != b[n/2])
        ++ans;             //如果n是奇数并且a[n/2]!=b[n/2],即a和b中间字符不同,ans加1;
    cout<<ans<<endl;
    return 0;
}