Codeforces Round #360 (Div. 2) B. Lovely Palindromes

B. Lovely Palindromes

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Pari has a friend who loves palindrome numbers. A palindrome number is a number that reads the same forward or backward. For example12321, 100001 and 1 are palindrome numbers, while 112 and 1021 are not.

Pari is trying to love them too, but only very special and gifted people can understand the beauty behind palindrome numbers. Pari loves integers with even length (i.e. the numbers with even number of digits), so she tries to see a lot of big palindrome numbers with even length (like a 2-digit 11 or 6-digit 122221), so maybe she could see something in them.

Now Pari asks you to write a program that gets a huge integer n from the input and tells what is the n-th even-length positive palindrome number?

Input

The only line of the input contains a single integer n (1 ≤ n ≤ 10^100 000).

Output

Print the n-th even-length palindrome number.

Examples

Input

1

Output

11

Input

10

Output

1001

Note

The first 10 even-length palindrome numbers are 11, 22, 33, ... , 88, 99 and 1001.

题意：求第n个回文偶数长度的数

思路：列几个例子就可以看出来，直接正反输出就可以了

ac代码：

#include<stdio.h>#include<math.h>#include<string.h>#include<stack>#include<set>#include<queue>#include<vector>#include<iostream>#include<algorithm>#define MAXN 1010000#define LL long long#define ll __int64#define INF 0xfffffff#define mem(x) memset(x,0,sizeof(x))#define PI acos(-1)#define eps 1e-8using namespace std;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=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}//headchar str[MAXN];char str2[MAXN];int main(){    scanf("%s",str);    int len=strlen(str);    for(int i=0;i<len;i++)        str2[len-i-1]=str[i];    str2[len]='\0';    printf("%s%s\n",str,str2);    return 0;}