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 $s$ consisting of $n$ lowercase Latin letters. Polycarp wants to remove exactly $k$ characters ($k\le n$) from the string $s$. Polycarp uses the following algorithm $k$ 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 $k$ times, thus removing exactly $k$ characters.

Help Polycarp find the resulting string.

Input

The first line of input contains two integers $n$ and $k$ ($1\le k\le n\le 4\cdot {10}^5$) — the length of the string and the number of letters Polycarp will remove.

The second line contains the string $s$ consisting of $n$ lowercase Latin letters.

Output

Print the string that will be obtained from $s$ after Polycarp removes exactly $k$ letters using the above algorithm $k$ 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(" ");
    
    return 0;
}