Problem: Vus the [Cossack](https://en.wikipedia.org/wiki/Cossacks) holds a programming competition, in which n people participate. He decided to award them all with pens and notebooks. It is known that Vus has exactly m pens and k notebooks.

Determine whether the Cossack can reward all participants, giving each of them at least one pen and at least one notebook.

Input

The first line contains three integers n, m, and k (1 ≤ n, m, k ≤ 100) — the number of participants, the number of pens, and the number of notebooks respectively.

Output

Print "Yes" if it possible to reward all the participants. Otherwise, print "No".

You can print each letter in any case (upper or lower).

Examples

Input


5 8 6


Output


Yes


Input


3 9 3


Output


Yes


Input


8 5 20


Output


No

Note

In the first example, there are 5 participants. The Cossack has 8 pens and 6 notebooks. Therefore, he has enough pens and notebooks.

In the second example, there are 3 participants. The Cossack has 9 pens and 3 notebooks. He has more than enough pens but only the minimum needed number of notebooks.

In the third example, there are 8 participants but only 5 pens. Since the Cossack does not have enough pens, the answer is "No".

Solution (in PYTHON):

Solution:
"""                         
Beautiful is better than ugly.
    Explicit is better than implicit.
Simple is better than complex.
    Complex is better than complicated.
Flat is better than nested.
    Sparse is better than dense.
                    
    * Readability counts *

 // Author : raj1307 - Raj Singh
 // Date   : 28.06.19

"""

from __future__ import division, print_function
import os,sys
from io import BytesIO, IOBase

if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip
else:
    _str = str
    str = lambda x=b"": x if type(x) is bytes else _str(x).encode()

def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().strip().split(" "))
def li(): return list(mi())

def dmain():
    sys.setrecursionlimit(100000000)
    threading.stack_size(40960000)
    thread = threading.Thread(target=main)
    thread.start()
    
#from collections import deque, Counter, OrderedDict,defaultdict
#from heapq import nsmallest, nlargest, heapify,heappop ,heappush, heapreplace
#from math import ceil,floor,log,sqrt,factorial,pow,pi
#from bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right
#from decimal import *,threading

abc='abcdefghijklmnopqrstuvwxyz'
abd={'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
mod,MOD=1000000007,998244353
vow=['a','e','i','o','u']
dx,dy=[-1,1,0,0],[0,0,1,-1]
def getKey(item): return item[0] 
def sort2(l):return sorted(l, key=getKey)
def d2(n,m,num):return [[num for x in range(m)] for y in range(n)]
def isPowerOfTwo (x): return (x and (not(x & (x - 1))) )
def decimalToBinary(n): return bin(n).replace("0b","")
def ntl(n):return [int(i) for i in str(n)]

def powerMod(x,y,p):
    res = 1
    x %= p
    while y > 0:
        if y&1:
            res = (res*x)%p
        y = y>>1
        x = (x*x)%p1
    return res

def gcd(x, y):
    while y:
        x, y = y, x % y
    return x
    
def isPrime(n) : # Check Prime Number or not 
    if (n <= 1) : return False
    if (n <= 3) : return True
    if (n % 2 == 0 or n % 3 == 0) : return False
    i = 5
    while(i * i <= n) : 
        if (n % i == 0 or n % (i + 2) == 0) : 
            return False
        i = i + 6
    return True


# For getting input from input.txt file 
#sys.stdin = open('input.txt', 'r')  
  
# Printing the Output to output.txt file 
#sys.stdout = open('output.txt', 'w') 

def main():
    
    
    
    
    #for _ in range(ii()):
    
    n,a,b=mi()
    
    if n<=a and n<=b:
        print('YES')
    else:
        print('NO')
    
    
      
        
        
        
    
    
    
    
    
    
    
    
    
    
    
    
    
        
    
    
    
    
    
    
    
    
    
    




# region fastio

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        if self.buffer.tell():
            return self.buffer.read()
        return os.read(self._fd, os.fstat(self._fd).st_size)

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", b" "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", b"\n"))
    if kwargs.pop("flush", False):
        file.flush()


sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
input = lambda: sys.stdin.readline().rstrip(b"\r\n")

# endregion

if __name__ == "__main__":
    main()
    #dmain()