Codeforces_718A

A. Efim and Strange Grade

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Efim just received his grade for the last test. He studies in a special school and his grade can be equal to any positive decimal fraction. First he got disappointed, as he expected a way more pleasant result. Then, he developed a tricky plan. Each second, he can ask his teacher to round the grade at any place after the decimal point (also, he can ask to round to the nearest integer).

There are t seconds left till the end of the break, so Efim has to act fast. Help him find what is the maximum grade he can get in no more than t seconds. Note, that he can choose to not use all t seconds. Moreover, he can even choose to not round the grade at all.

In this problem, classic rounding rules are used: while rounding number to the n-th digit one has to take a look at the digit n + 1. If it is less than 5 than the n-th digit remain unchanged while all subsequent digits are replaced with 0. Otherwise, if the n + 1 digit is greater or equal to 5, the digit at the position n is increased by 1 (this might also change some other digits, if this one was equal to 9) and all subsequent digits are replaced with 0. At the end, all trailing zeroes are thrown away.

For example, if the number 1.14 is rounded to the first decimal place, the result is 1.1, while if we round 1.5 to the nearest integer, the result is 2. Rounding number 1.299996121 in the fifth decimal place will result in number 1.3.

Input

The first line of the input contains two integers n and t (1 ≤ n ≤ 200 000, 1 ≤ t ≤ 109) — the length of Efim's grade and the number of seconds till the end of the break respectively.

The second line contains the grade itself. It's guaranteed that the grade is a positive number, containing at least one digit after the decimal points, and it's representation doesn't finish with 0.

Output

Print the maximum grade that Efim can get in t seconds. Do not print trailing zeroes.

Examples

input

6 1
10.245

output

10.25

input

6 2
10.245

output

10.3

input

3 100
9.2

output

9.2

Note

In the first two samples Efim initially has grade 10.245.

During the first second Efim can obtain grade 10.25, and then 10.3 during the next second. Note, that the answer 10.30 will be considered incorrect.

In the third sample the optimal strategy is to not perform any rounding at all.

题意：给一个小数点后至少有一位的小数（200000），t次四舍五入，只能舍小数点后的数，找最大的。1e9次查询。

模拟题。

之前每次都从小数点开始找第一个大于5的数，最开始一直超时。。。

其实，最开始找一次离小数点最近的大于5的数的位置。

循环中每次进行四舍五入后，从之前找到的那位往前找一个大于5的数，这个数一定是离小数点最近的大于5的数。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
#include<cmath>
#include<string>

using namespace std;

char str[200005];
int main()
{
    int l,t;

    str[0]='0';
    scanf("%d%d%s",&l,&t,str+1);

    //cout<<str<<endl;
    int len=strlen(str)-1,tail=len,point;
    for(int i=0; i<=len; i++)
        if(str[i]=='.')
            point=i;
    int loc=-1;
    for(int i=len; i>point; i--)
    {
        if(str[i]>='5')
        {
            loc=i;
        }
    }
    if(loc>0)
    {
        tail=loc;
        while(t--)
        {
            len=tail;
            loc=len;
            // cout<<"len:"<<len<<endl;
            //cout<<len<<"*"<<endl;
            if(point>tail)
                break;
            for(int i=len; i>point; i--)
            {
                if(str[i]>='5')
                {
                    loc=i;
                    break;
                }
            }

            int jin=0;
            if(str[loc]>='5')
                jin=1;
            if(loc==len&&jin==0)
                break;
            //cout<<loc<<"*"<<jin<<endl;
            tail=loc-1;
            if(jin>0&&str[tail]+jin<='9'&&str[tail]!='.')
            {
                str[tail]+=jin;
                jin=0;
            }
            //cout<<str<<"*"<<endl;
            while(jin>0)
            {
                if(str[tail]=='.')
                {
                    tail--;
                }
                str[tail]+=jin;
                if(str[tail]<='9')
                    jin=0;
                else
                {
                    str[tail]-=10;
                    tail--;
                }
            }
        }
    }
    //cout<<str[]<<endl;
    if(str[0]=='0')
    {
        if(tail<point)
            tail=point;
        for(int i=1; i<=tail; i++)
        {
            if(i==tail&&str[tail]=='.')
                continue;
            putchar(str[i]);
        }
        putchar('
');
    }
    else
    {
        if(tail<point)
            tail=point;
        for(int i=0; i<=tail; i++)
        {
            if(i==tail&&str[tail]=='.')
                continue;
            putchar(str[i]);
        }
        putchar('
');
    }
    //cout<<str<<endl;

    return 0;
}