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
p=input().rstrip().split(' ')
s=input().rstrip()
x=list(s)
k=int(p[1])
l=['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']
'''for i in range(0,k):
    if len(x)==0:
        break;
    else:
        for j in range(0,len(l)):
            if l[j] in x:
                V=x.index(l[j])
                del(x[V])
                break;
if len(x)!=0:
    print(''.join(x))'''

q=[]
for i in range(0,len(l)):
    q.append(x.count(l[i]))
for i in range(0,len(q)):
    if int(q[i])<k:
        k=k-int(q[i])
        q[i]=0;
    elif int(q[i])==k:
        q[i]=0;
        k=0;
        break;
    elif int(q[i])>k:
        q[i]=int(q[i])-k;
        break;
D=[]
for i in range(len(x)-1,-1,-1):
    if q[ord(x[i])-97]>0:
        D.append(x[i])
        q[ord(x[i])-97]-=1;
if len(D)!=0:
    D.reverse();
    print(''.join(D))
