Two Strings Swaps CodeForces - 1006D

问题:

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 1 to n, inclusive.

You are allowed to do the following changes:

• Choose any index ii (1≤i≤n) and swap characters ai and bi;
• Choose any index ii (1≤i≤n) and swap characters ai and an−i+1
• Choose any index ii (1≤i≤n) and swap characters bi and bn−i+1

Note that if n is odd, you are formally allowed to swap a⌈n2⌉ with 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≤n), any character cc and set ai:=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 moves to 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≤105) — the length of strings aa and bb.

The second line contains the string a consisting of exactly n lowercase English letters.

The third line contains the string b consisting of exactly n 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 b with a sequence of changes from the list above.

Examples

Input

7
abacaba
bacabaa

Output

4

Input

5
zcabd
dbacz

Output

0

Note

In the first example preprocess moves are as follows: a1:='b', a3:='c', a4:='a' and a5:='b'. Afterwards, a="bbcabba". Then we can obtain equal strings by the following sequence of changes: swap(a2,b2) and swap(a2,a6). There is no way to use fewer than 4 preprocess moves before a sequence of changes to make string equal, so the answer in this example is 4.

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(a2,a4).

大意：

给你两个等长的字符串啊a，b，问你最少要改变a中多少个字符，然后通过任意次上述三种变换，使两字符串相等。

思路：

去找a[i],a[n-1-i]和b[i],b[n-1-i]的关系，（1）如果四个都一样就不用换，（2）如果三个一样，只需要换一个，（3）如果两两一样就不用变，（4）只有两个一样，也要看情况，（5）如果四个都不一样就要换两个。

代码：

#include<map>
#define N 100100
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char a[N],b[N];
int n,sum;
int main()
{
    while(~scanf("%d",&n))
    {
        scanf("%s %s",a,b);
        sum=0;
        for (int i=0; i<n/2; i++)
        {
            map<char,int>m;
            m[a[i]]++;
            m[a[n-i-1]]++;
            m[b[i]]++;
            m[b[n-i-1]]++;
            if (m.size()==4)//四个都不一样
                sum+=2;
            else if(m.size()==3)//只有两个一样
            {
                sum+=1;
                if(a[i]==a[n-i-1])//a组的两个相等，就说明和b中的不等要改变两个
                    sum+=1;
            }
            else if(m.size()==2)//属于（2）（3）情况，
            {
                if(m[a[i]]!=2)//属于（2）
                    sum+=1;
            }
        }
        if (n%2&&a[n/2]!=b[n/2])//如果字符长为奇数要特判
            sum++;
        printf("%d\n",sum);
    }
    return 0;
}