codeforces——313A——Ilya and Bank Account

A. Ilya and Bank Account

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Ilya is a very clever lion, he lives in an unusual city ZooVille. In this city all the animals have their rights and obligations. Moreover, they even have their own bank accounts. The state of a bank account is an integer. The state of a bank account can be a negative number. This means that the owner of the account owes the bank money.

Ilya the Lion has recently had a birthday, so he got a lot of gifts. One of them (the gift of the main ZooVille bank) is the opportunity to delete the last digit or the digit before last from the state of his bank account no more than once. For example, if the state of Ilya's bank account is -123, then Ilya can delete the last digit and get his account balance equal to -12, also he can remove its digit before last and get the account balance equal to -13. Of course, Ilya is permitted not to use the opportunity to delete a digit from the balance.

Ilya is not very good at math, and that's why he asks you to help him maximize his bank account. Find the maximum state of the bank account that can be obtained using the bank's gift.

Input

The single line contains integer n (10 ≤ |n| ≤ 10⁹) — the state of Ilya's bank account.

Output

In a single line print an integer — the maximum state of the bank account that Ilya can get.

Examples

Input

2230

Output

2230

Input

-10

Output

0

Input

-100003

Output

-10000

Note

In the first test sample Ilya doesn't profit from using the present.

In the second test sample you can delete digit 1 and get the state of the account equal to 0.

可以删掉最后一个数或者倒数第二个数也可以不删，问在哪种情况资产最大

水，写的有点长，本来想着往数组里套，以简短代码，后来想着没必要就没有改

#include<iostream>#include<algorithm>#include<stdio.h>#include<cstring>using namespace std;int main(){    int n;    while(cin>>n!=NULL)    {        if(n<0)        {            char s[20];            sprintf(s,"%d",n);            int n2=strlen(s);            s[n2-1]='\0';            int x1;            sscanf(s,"%d",&x1);            sprintf(s,"%d",n);            s[n2-2]=s[n2-1];            s[n2-1]='\0';            int x2;            sscanf(s,"%d",&x2);            n=(x1>x2?x1:x2);        }        cout<<n<<endl;    }    return 0;}