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 PYTHON3):

Solution:
#########################################################################################################\
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###################################The_Apurv_Rathore#####################################################
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import sys,os,io
from sys import stdin
from math import log, gcd, ceil
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop
from bisect import bisect_left , bisect_right
import math 

sys.setrecursionlimit(100000)

alphabets = list('abcdefghijklmnopqrstuvwxyz')


def isPrime(x):
    for i in range(2,x):
        if i*i>x:
            break
        if (x%i==0):
            return False
    return True
def ncr(n, r, p):  
    num = den = 1
    for i in range(r):
        num = (num * (n - i)) % p
        den = (den * (i + 1)) % p
    return (num * pow(den,p - 2, p)) % p
def primeFactors(n): 
    l = []
    while n % 2 == 0: 
        l.append(2)
        n = n / 2
    for i in range(3,int(math.sqrt(n))+1,2): 
        while n % i== 0: 
            l.append(int(i))
            n = n / i 
    if n > 2: 
        l.append(n)
    c = dict(Counter(l))
    return list(set(l))
    # return c

def power(x, y, p) : 
	res = 1
	x = x % p 
	if (x == 0) : 
		return 0
	while (y > 0) : 
		if ((y & 1) == 1) : 
			res = (res * x) % p 
		y = y >> 1	 # y = y/2 
		x = (x * x) % p 		
	return res 

#____________________GetPrimeFactors in log(n)________________________________________
def sieveForSmallestPrimeFactor():
    MAXN = 100001
    spf = [0 for i in range(MAXN)]
    spf[1] = 1
    for i in range(2, MAXN):
        spf[i] = i
    for i in range(4, MAXN, 2):
        spf[i] = 2
    for i in range(3, math.ceil(math.sqrt(MAXN))):
        if (spf[i] == i):
            for j in range(i * i, MAXN, i): 
                if (spf[j] == j):
                    spf[j] = i
    return spf
def getPrimeFactorizationLOGN(x):
    spf = sieveForSmallestPrimeFactor()
    ret = list()
    while (x != 1):
        ret.append(spf[x])
        x = x // spf[x]  
    return ret
#____________________________________________________________



def SieveOfEratosthenes(n): 
    #time complexity = nlog(log(n))
    prime = [True for i in range(n+1)]
    p = 2
    pr = []
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n+1, p):
                prime[i] = False
        p += 1
    for i in range(2,len(prime)):
        if prime[i]:
            pr.append(i)
    return pr
def si():
    return input()
def divideCeil(n,x):
    #return (n+1)//x
    if (n%x==0):
        return n//x
    return n//x+1
def ii():
    return int(input())
def li():
    return list(map(int,input().split()))

#__________________________TEMPLATE__________________OVER_______________________________________________________


if(os.path.exists('input.txt')):
    sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w") 
else:
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline


def solve():
    n,m,k = li()
    if n<=m and n<=k:
        print("YES")
    else:
        print("NO")

t = 1
# t = ii()
for _ in range(t):
    solve()