You are given a string s consisting of n lowercase Latin letters. Polycarp wants to remove exactly k characters (k ≤ 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 ≤ k ≤ n ≤ 4 ⋅ 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

15 3
cccaabababaccbc


Output

cccbbabaccbc


Input

15 9
cccaabababaccbc


Output

cccccc


Input

1 1
u


Output
let=['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z']
cont=[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]

q = input().split()
tam,num = int(q[0]), int(q[1])
pala=list(input())


for i in pala:
    cont[let.index(i)]+=1
#sei quantas de cada tem.

for j in range(len(cont)):
    if num>=cont[j]:
        num=num-cont[j]
        cont[j]=0
    else:
        cont[j]=cont[j]-num
        num=0
        break
#apaguei todas que precisam.


#falta imprimir a string nova
palaN=''
for k in range(len(pala)-1,-1,-1): #tras para frente
    if cont[let.index(pala[k])] > 0:
        palaN+=pala[k]
        cont[let.index(pala[k])]-=1
        
print(palaN[::-1]) #inverte novamente

  	   				  	 			  	        	 	