Problem: A word or a sentence in some language is called a pangram if all the characters of the alphabet of this language appear in it at least once. Pangrams are often used to demonstrate fonts in printing or test the output devices.

You are given a string consisting of lowercase and uppercase Latin letters. Check whether this string is a pangram. We say that the string contains a letter of the Latin alphabet if this letter occurs in the string in uppercase or lowercase.

Input

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

The second line contains the string. The string consists only of uppercase and lowercase Latin letters.

Output

Output "YES", if the string is a pangram and "NO" otherwise.

Examples

Input

12
toosmallword


Output

NO


Input

35
TheQuickBrownFoxJumpsOverTheLazyDog


Output

YES

Solution (in PYTHON3):

Solution:
import sys
def rr(): return sys.stdin.readline().rstrip()
#def pp(*args, end="\n"): sys.stdout.write(" ".join(map(str, args))+end)
rp = print
def print(*args, **kwargs): rp("DBG:# ", *args, **kwargs, file=sys.stderr)
def rri(): return int(rr())
def rrl(): return list(map(int, rr().split()))
def rrt(): return tuple(map(int, rr().split()))
def rrle(): return (line.rstrip() for line in sys.stdin.readlines())
def rrtl(): return [tuple(map(int, l.split())) for l in rrle()]
from collections import defaultdict
def mus(d=lambda: 0): return defaultdict(lambda: defaultdict(d))
def dd0(d=lambda: 0): return defaultdict(d)
def ms(x, y, d=0): return [[0]*y for i in range(x)]
def ar(x, d=0): return [d]*x
def ppm(m, n=0, x=0, y=0): print("\n".join(("\t".join((str(m[j][i]) for j in range(y or n))) for i in range(x or n))))
def ppa(a, n): print("\t".join(map(str, a[0:n])))
def ppl(a=0, x=[1]): x[0]=a or (x[0]+1); print("-"+"- -"*14+"-# "+str(x[0]-2)+" #-"+"- -"*14+"-")
def fltn(a): (i for sub in a for i in sub)
def yn(x): return "YES" if x else "NO"
def cpc(arr): return zip(arr, arr[1:])
from itertools import accumulate
from itertools import groupby
from itertools import permutations
def grp(x, key=None): return ((i, sum(1 for _ in g)) for i, g in groupby(x, key=key))
import math
def rnar(): return (*rrl(), rrl())
def rn(): return (*rrl(),)
def dpp(a, b=""): print("#| {} |#:^ {} ^:".format(a, b))
def str_join(i, sep=" "): return sep.join(map(str, i))
MOD17 = 10**9 + 7

def read():
    rri()
    return (rr(),)

def solve(s):
    ans = yn(len(set(s.lower()))==26)
    return ans

if __name__ == "__main__":
    #test_count = rri()
    test_count = 1
    for _ in range(test_count):
        input_data = read()
        result = solve(*input_data)
        rp(result)