The recent All-Berland Olympiad in Informatics featured n participants with each scoring a certain amount of points.

As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria: 

  * At least one participant should get a diploma. 
  * None of those with score equal to zero should get awarded. 
  * When someone is awarded, all participants with score not less than his score should also be awarded. 



Determine the number of ways to choose a subset of participants that will receive the diplomas.

Input

The first line contains a single integer n (1 ≤ n ≤ 100) — the number of participants.

The next line contains a sequence of n integers a1, a2, ..., an (0 ≤ ai ≤ 600) — participants' scores.

It's guaranteed that at least one participant has non-zero score.

Output

Print a single integer — the desired number of ways.

Examples

Input

4
1 3 3 2


Output

3


Input

3
1 1 1


Output

1


Input

4
42 0 0 42


Output

1

Note

There are three ways to choose a subset in sample case one.

  1. Only participants with 3 points will get diplomas. 
  2. Participants with 2 or 3 points will get diplomas. 
  3. Everyone will get a diploma! 



The only option in sample case two is to award everyone.

Note that in sample case three participants with zero scores cannot get anything.
#import resource
#import sys
#resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])
#import threading
#threading.stack_size(2**26)
#sys.setrecursionlimit(0x1000000)
from sys import stdin, stdout
mod=(10**9)+7
mod1=mod-1
def modinv(n,p):
    return pow(n,p-2,p)
fact=[1]
for i in range(1,100001):
    fact.append((fact[-1]*i)%mod)
def ncr(n,r,p):
    t=((fact[n])*(modinv(fact[r],p)%p)*(modinv(fact[n-r],p)%p))%p
    return t
def GCD(x, y):
   while(y):
       x, y = y, x % y
   return x
def BS(arr, l, r, x):
    if r >= l:
        mid = l + (r - l)/2
        if arr[mid] == x:
            return mid
        elif arr[mid] > x:
            return BS(arr, l, mid-1, x)
        else:
            return BS(arr, mid+1, r, x)
    else:n -1
from bisect import bisect_left as bl
from bisect import bisect_right as br
import itertools
import math
import heapq
from random import randint as rn
from Queue import Queue as Q
def comp(x,y):
    if(x[0]<y[0]):
        return -1
    elif(x[0]==y[0]):
        if(x[1]<y[1]):
            return -1
        else:
            return 1
    else:
        return 1
def find(x):
    if(p[x]!=x):
        p[x]=find(p[x])
    return p[x]
def union(x,y):
    m=find(x)
    n=find(y)
    if(m!=n):
        s[0]-=1
        if(r[m]>r[n]):
            p[n]=m
        elif(r[n]>r[m]):
            p[m]=n
        else:
            p[n]=m
            r[m]+=1
"""*********************************************************************************"""
n=input()
w=map(int,raw_input().split())
a=set()
for i in range(n):
    if(w[i]!=0):
        a.add(w[i])
print len(a)
