Codeforces Round #373 (Div. 2) C： Efim and Strange Grade



C. 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 thant seconds. Note, that he can choose to not use allt 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 then-th digit one has to take a look at the digitn + 1. If it is less than 5 than the n-th digit remain unchanged while all subsequent digits are replaced with0. Otherwise, if the n + 1 digit is greater or equal to5, the digit at the position n is increased by 1 (this might also change some other digits, if this one was equal to9) 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 is2. Rounding number 1.299996121 in the fifth decimal place will result in number1.3.

Input

The first line of the input contains two integersn 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 with0.

Output

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

Examples

Input

6 110.245

Output

10.25

Input

6 210.245

Output

10.3

Input

3 1009.2

Output

9.2

Note

In the first two samples Efim initially has grade10.245.

During the first second Efim can obtain grade 10.25, and then 10.3 during the next second. Note, that the answer10.30 will be considered incorrect.

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

题意：给你一个小数和小数点后面t次进位的机会，问最后能变成多大的数。

题解：看到t很大，就没敢写， 后来发现就是一个贪心，每次我们从距离小数点最近的并且能进位的贪，然后将此位后面的删除。

代码：

#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<iostream>#include<algorithm>#include<set>using namespace std;char s[200005];set<int>st;int main(){    int n,t,d;    scanf("%d%d%s",&n,&t,s);    for(int i=0;i<n;i++)        if(s[i]=='.')d=i;    for(int i=d+1;i<n;i++)        if(s[i]>='5')st.insert(i);    int now=n;    while(t--)    {        if(st.empty() || *st.begin()>=now || *st.begin()<d)break;        int u=*st.begin();        if(s[u-1]=='.')u--;        if(++s[u-1]>='5')st.insert(u-1);        now=u;    }    bool flag=0;    for(int i=d-1;i>=0;i--)    {        if(i)s[i-1]+=(s[i]-'0')/10;        else if(s[i]>'9')flag=1;        s[i]=(s[i]-'0')%10+'0';    }    if(flag)printf("1");    for(int i=0;i<now;i++)        printf("%c",s[i]);    return 0;}