CoderForces999C-Alphabetic Removals

C. Alphabetic Removals

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a string ss consisting of nn lowercase Latin letters. Polycarp wants to remove exactly kk characters (k≤nk≤n) from the string ss. Polycarp uses the following algorithm kk times:

• if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
• if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
• ...
• remove the leftmost occurrence of the letter 'z' and stop the algorithm.

This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly kk times, thus removing exactly kk characters.

Help Polycarp find the resulting string.

Input

The first line of input contains two integers nn and kk (1≤k≤n≤4⋅1051≤k≤n≤4⋅105) — the length of the string and the number of letters Polycarp will remove.

The second line contains the string ss consisting of nn lowercase Latin letters.

Output

Print the string that will be obtained from ss after Polycarp removes exactly kk letters using the above algorithm kk times.

If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).

Examples

input

Copy

15 3
cccaabababaccbc

output

Copy

cccbbabaccbc

input

Copy

15 9
cccaabababaccbc

output

Copy

cccccc

input

Copy

1 1
u

output

Copy

AC代码为：

#include<bits/stdc++.h>
using namespace std;
const int maxn=4e5+10;
struct Node{
    int id,temp;
    char c;
} node[maxn];
bool cmp1(Node a,Node b)
{
    return a.c==b.c? a.id<b.id : a.c-'a'<b.c-'a';
}
bool cmp2(Node a,Node b)
{
    return a.id<b.id;
}
int n,k;
char s1[maxn];
int main()
{
    scanf("%d%d",&n,&k);
    scanf("%s",s1);
    int len=strlen(s1);
    for(int i=0;i<len;i++)
    {
        node[i].c=s1[i];
        node[i].id=i;
        node[i].temp=0;
    }
    sort(node,node+len,cmp1);
    for(int i=0;i<k;i++) node[i].temp=1;
    sort(node,node+len,cmp2);
    for(int i=0;i<len;i++)
    {
        if(node[i].temp==1) continue;
        else printf("%c",node[i].c); 
    }
    printf("\n");
    
    return 0;
}