Mishka has got n empty boxes. For every i (1 ≤ i ≤ n), i-th box is a cube with side length ai.

Mishka can put a box i into another box j if the following conditions are met:

  * i-th box is not put into another box; 
  * j-th box doesn't contain any other boxes; 
  * box i is smaller than box j (ai < aj). 



Mishka can put boxes into each other an arbitrary number of times. He wants to minimize the number of visible boxes. A box is called visible iff it is not put into some another box.

Help Mishka to determine the minimum possible number of visible boxes!

Input

The first line contains one integer n (1 ≤ n ≤ 5000) — the number of boxes Mishka has got.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the side length of i-th box.

Output

Print the minimum possible number of visible boxes.

Examples

Input

3
1 2 3


Output

1


Input

4
4 2 4 3


Output

2

Note

In the first example it is possible to put box 1 into box 2, and 2 into 3.

In the second example Mishka can put box 2 into box 3, and box 4 into box 1.
#edu round 34, C
#cubic boxes, box i can be packed inside j if side ai<aj
#find minimum visible boxes
# 1 ≤ n ≤ 5000, 1 ≤ ai ≤ 10^9
# strategy:
# sort all boxes by side
# find a smaller number in answer list and replace with next box if possible
# else, add box to answer list if it is not bigger
# sort answer list to improve performance
def packing(a):
    a.sort()
    ans=[]
    for i in range(len(a)):
        found=False
        if len(ans)>0 and ans[0]<a[i]:  # check first in answer as it is sorted
                ans[0]=a[i]  # replace answer
                ans.sort()   # sort again
        else:
            ans.append(a[i]) # add new box that cannot pack smaller ones
            ans.sort()       # sort to improve performance
    return len(ans)

# better idea from editorial
# find the max boxes of the same size, that should determine the result
def packing2(a):
    a.sort()
    ans=1  #assume at least one box
    last=a[0]
    lastIndex=0
    for i in range(1,len(a)):
        if a[i]==last:
            continue
        count=i-lastIndex  #find next size
        if count>ans:
            ans=count
        last=a[i]
        lastIndex=i
    count=len(a)-lastIndex  #check last box too
    if count>ans:
        ans=count
    return ans
    
#print (packing2([1,2,3])==1)
#print (packing2([4,2,4,3,5,6,7,9,9,9,7,6,3,1,8,8])==3)


def ni():
    s=input()
    while len(s)==0:
        s=input()
    try:
        return int(s)
    except:
        return 0

def nia():
    s=input()
    while len(s)==0:
        s=input()
    s=s.split()
    iVal=[];
    for i in range (len(s)):
        iVal.append(int(s[i]))
    return iVal

n=ni()
print (packing2(nia()))
