Two Strings Swaps
You are given two strings aa and bb consisting of lowercase English letters, both of length n. The characters of both strings have indices from 1 to n, inclusive.
You are allowed to do the following changes:
- Choose any index i (1≤i≤n) and swap characters ai and bi;
- Choose any index i (1≤i≤n) and swap characters ai and an−i+1;
- Choose any index i (1≤i≤n) and swap characters bi and bn−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 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≤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
7 abacaba bacabaa
Output
4
Input
5 zcabd dbacz
Output
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 44 preprocess 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).
emmm... 我打的第一場cf上的原題，結果昨天我根本就沒看這道題....
題意：有兩個字串a，b，可以進行 a[i]和a[n-i-1]交換 或 b[i]和b[n-i-1]交換 或a[i]和b[i]交換。求剛開始最少對 a進行修改多少個字元才能使兩個字串可以通過上述三種交換方式得到兩個相同的字串。
三種交換方式：
思路：根據三種交換方式我們可以知道，這四個字元是可以在這四個位置任意交換的。那麼最少的預處理次數就是要看每對（四個字元）裡面的字元匹配情況了。
情況一：四個字元都相等，那麼就不需要進行操作就可以達到目標狀態（a，b字串上下位置對應相等）。
情況二：四個字元中有兩種字元。
1.如果是（2，2）的話，是可以通過交換轉換達到目的狀態的；
2.如果是（1，3）的話，肯定需要改變任意一個字元，達到（2，2）或者是 情況一；
情況三：四個字元中有三種字元。
1.如果是a[i]==a[n-i-1],那麼就可以推出 b[i]!=b[n-i-1]!=(a[i]==a[n-i-1])，又因為只能改變a，那麼是必須要改變a中的兩個字 符的（先改變，後交換）。
2.反之，只需要改變一個即可。
情況四：四個字元又四種字元。因此改變任意兩個即可。
情況五：整個字串的長度是奇數，那麼中間就肯定剩下兩個字元沒有處理，如果這兩個字元不相等，就必須改變一個了。
程式碼如下：
#include<cstdio>
#include<cstring>
#include<map>
#include<algorithm>
using namespace std;
char s[100010],t[100010];
int main()
{
int n,i,j;
scanf("%d",&n);
scanf("%s%s",s,t);
int ans=0;
for(i=0; i<n/2; i++)
{
map<char,int>v; //判斷有幾種字元，並記錄各自個數
v[s[i]]++;
v[t[i]]++;
v[s[n-i-1]]++;
v[t[n-i-1]]++;
if(v.size()>=2)
{
if(v.size()==2) //有兩種字元
{
if(v[s[i]]!=2) //（1，3）情況
ans++;
}
else if(v.size()==3) //有三種字元
{
if(s[i]==s[n-i-1]) // 因為是先改變再交換，且只能改變s字串，所以這種情況下要改變兩個
ans+=2;
else //反之，改變一個
ans++;
}
else //有四種字元
ans+=2;
}
}
if(n%2) //奇數
if(s[n/2]!=t[n/2])//判斷中間的兩個字元是否相等
ans++;
printf("%d\n",ans);
}