You are given two strings ss and tt, both consisting of exactly kk lowercase Latin letters, ss is lexicographically less than tt.
Let's consider list of all strings consisting of exactly kk lowercase Latin letters, lexicographically not less than ss and not greater than tt (including ss and tt) in lexicographical order. For example, for k=2k=2, s=s="az" and t=t="bf" the list will be ["az", "ba", "bb", "bc", "bd", "be", "bf"].
Your task is to print the median (the middle element) of this list. For the example above this will be "bc".
It is guaranteed that there is an odd number of strings lexicographically not less than ss and not greater than tt.
Input
The first line of the input contains one integer kk (1≤k≤2⋅1051≤k≤2⋅105) — the length of strings.
The second line of the input contains one string ss consisting of exactly kk lowercase Latin letters.
The third line of the input contains one string tt consisting of exactly kk lowercase Latin letters.
It is guaranteed that ss is lexicographically less than tt.
It is guaranteed that there is an odd number of strings lexicographically not less than ss and not greater than tt.
Output
Print one string consisting exactly of kk lowercase Latin letters — the median (the middle element) of list of strings of length kk lexicographically not less than ss and not greater than tt.
Examples
Input
2
az
bf
Output
bc
Input
5
afogk
asdji
Output
alvuw
Input
6
nijfvj
tvqhwp
Output
qoztvz 题意：给你两个只含有小写字母的字符串， s和t，保证s的字典序比t小。把s和t看成一个26进制的数，让你输出这两个数的中位数。 思路：首先把两个字符串的每一个位上的数值加起来，然后向前进位。
然后从头开始遍历加起来的那个数值，如果数值为偶数，那么中位数的这一位就是数值/2，如果是奇数，那么中位数的这一位是/2且向下取整。
并把那个多余的一加到下一位中，（注意上一位的1到下一位是26），最后输出答案即可。 *（主要是通过咱们熟悉的十进制来找到转换的规律，然后处理。）
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d\n",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=;while(b){if(b%)ans=ans*a%MOD;a=a*a%MOD;b/=;}return ans;}
inline void getInt(int* p);
const int maxn=;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int a[maxn];
int b[maxn];
int main()
{
// freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
// freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
gbtb;
// repd(i,0,25)
// {
// cout<<(char)('a'+i)<<" ";
// }
string xia;
string shang;
int k;
cin>>k>>xia>>shang;
for(int i=k-;i>=;i--)
{
// a[i-1]+=(shang[i]+xia[i]-'a'-'a')/26;
a[i]+=(shang[i]+xia[i]-'a'-'a');
}
for(int i=k-;i>;i--)
{
a[i-]+=a[i]/;
a[i]%=;
}
// repd(i,0,k)
// {
// db(a[i]);
// }
for (int i = ; i < k; ++i)
{
if(a[i]&)
{
a[i+]+=;;
b[i]=a[i]/;
}else
{
b[i]=a[i]/;
}
}
for (int i = ; i < k; ++i)
{
cout<<(char)(b[i]+'a');
/* code */
}
cout<<endl;
return ;
} inline void getInt(int* p) {
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '\n');
if (ch == '-') {
*p = -(getchar() - '');
while ((ch = getchar()) >= '' && ch <= '') {
*p = *p * - ch + '';
}
}
else {
*p = ch - '';
while ((ch = getchar()) >= '' && ch <= '') {
*p = *p * + ch - '';
}
}
}
