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:
# -*- coding: utf-8 -*-
"""
Created on Wed Jun 26 18:50:58 2019

@author: Hamadeh
"""

class cinn:
    def __init__(self):
        self.x=[]
    def cin(self,t=int):
        if(len(self.x)==0):
            a=input()
            self.x=a.split()
            self.x.reverse()
        return self.get(t)
    def get(self,t):
        return t(self.x.pop())
    def clist(self,n,t=int): #n is number of inputs, t is type to be casted
        l=[0]*n
        for i in range(n):
            l[i]=self.cin(t)
        return l
    def clist2(self,n,t1=int,t2=int,t3=int,tn=2):
        l=[0]*n
        for i in range(n):
            if(tn==2):
                a1=self.cin(t1)
                a2=self.cin(t2)
                l[i]=(a1,a2)
            elif (tn==3):
                a1=self.cin(t1)
                a2=self.cin(t2)
                a3=self.cin(t3)
                l[i]=(a1,a2,a3)
        return l
    def clist3(self,n,t1=int,t2=int,t3=int):
        return self.clist2(self,n,t1,t2,t3,3)
    def cout(self,i,ans=''):   
        if(ans==''):
            print("Case #"+str(i+1)+":", end=' ')
        else:
            print("Case #"+str(i+1)+":",ans)
    def printf(self,thing):
        print(thing,end='')
    def countlist(self,l,s=0,e=None):
        if(e==None):
            e=len(l)
        dic={}
        for el in range(s,e):
            if l[el] not in dic:
                dic[l[el]]=1
            else:
                dic[l[el]]+=1
        return dic
    def talk (self,x):
        print(x,flush=True)
    def dp1(self,k):
        L=[-1]*(k)
        return L
    def dp2(self,k,kk):
        L=[-1]*(k)
        for i in range(k):
            L[i]=[-1]*kk
        return L
    def isprime(self,n):
        if(n==1 or n==0):
            return False
        for i in range(2,int(n**0.5+1)):
            if(n%i==0):
                return False
        return True
    def factors(self,n):    
        from functools import reduce
        return set(reduce(list.__add__, 
                    ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
    def nthprime(self,n):
        #usable up to 10 thousand
        i=0
        s=2
        L=[]
        while(i<n):
            while(not self.isprime(s)):
                s+=1
            L.append(s)
            s+=1
            i+=1
        return L
    def matrixin(self,m,n,t=int):
        L=[]
        for i in range(m):
            p=self.clist(n,t)
            L.append(p)
        return L
    def seive(self,k):
        #1000000 tops
        n=k+1
        L=[True]*n
        L[1]=False
        L[0]=False
        for i in range(2,n):
            if(L[i]==True):
                for j in range(2*i,n,i):
                    L[j]=False
        return L
    def seiven(self,n,L):
        i=0
        for j in range(len(L)):
            if(L[j]==True):
                i+=1
            if(i==n):
                return j
    def matrixin2(self,m,t=int):
        L=[]
        for i in range(m):
            iny=self.cin(str)
            lsmall=[]
            for el in iny:
                lsmall.append(t(el))
            L.append(lsmall)
        return L
c=cinn()
n=c.cin()
x=c.cin()
y=c.cin()
if(n<=x and n<=y):
    print("Yes")
else:
    print("No")