human
Hereever handle is chosen, I must be the first, then tourist and Petr must be the last given the following:
A way to make a new task is to make it nondeterministic or probabilistic. For example, the hard task of Topcoder SRM 595, Constellation, is the probabilistic version of a convex hull.

Let's try to make a new task. Firstly we will use the following task. There are n people, sort them by their name. It is just an ordinary sorting problem, but we can make it more interesting by adding nondeterministic element. There are n people, each person will use either his/her first name or last name as a handle. Can the lexicographical order of the handles be exactly equal to the given permutation p?

More formally, if we denote the handle of the i-th person as hi, then the following condition must hold: .

Input

The first line contains an integer n (1 ≤ n ≤ 105) — the number of people.

The next n lines each contains two strings. The i-th line contains strings fi and si (1 ≤ |fi|, |si| ≤ 50) — the first name and last name of the i-th person. Each string consists only of lowercase English letters. All of the given 2n strings will be distinct.

The next line contains n distinct integers: p1, p2, ..., pn (1 ≤ pi ≤ n).

Output

If it is possible, output "YES", otherwise output "NO".

Examples

Input

3
gennady korotkevich
petr mitrichev
gaoyuan chen
1 2 3

Output

NO

Input

3
gennady korotkevich
petr mitrichev
gaoyuan chen
3 1 2

Output

YES

Input

2
galileo galilei
nicolaus copernicus
2 1

Output

YES

Input

10
rean schwarzer
fei claussell
alisa reinford
eliot craig
laura arseid
jusis albarea
machias regnitz
sara valestin
emma millstein
gaius worzel
1 2 3 4 5 6 7 8 9 10

Output

NO

Input

10
rean schwarzer
fei claussell
alisa reinford
eliot craig
laura arseid
jusis albarea
machias regnitz
sara valestin
emma millstein
gaius worzel
2 4 9 6 5 7 1 3 8 10

Output

YES

Note

In example 1 and 2, we have 3 people: tourist, Petr and me (cgy4ever). You can see that whatever handle is chosen, I must be the first, then tourist and Petr must be the last.

In example 3, if Copernicus uses "copernicus" as his handle,...
assistant
#Hereever handle is chosen, I must be the first, then tourist and Petr must be the last.

def ar():
    return map(int,raw_input().split())
n=int(raw_input())
a=[]
class A:
    def __init__(self,f,l,p):
        self.f=f
        self.l=l
        self.p=p
    def __lt__(self,r):
        return self.p<r.p

for i in range(n):
    f,l=sorted(raw_input().split())
    a.append([f,l])
    
s=ar()
obs=[]

for i in range(n):
    obs.append(A(a[s[i]-1][0],a[s[i]-1][1],i))

##for i in obs:
##    print i.f,i.l,i.p

obs=sorted(obs)
##
##for i in obs:
##    print i.f,i.l,i.p

prev=obs[0].f
flag=True
for i in range(1,n):
    if obs[i].f>prev:
        prev=obs[i].f
    elif obs[i].l>prev:
        prev=obs[i].l
    else:
        flag=False
        break
if flag:
    print"YES"
else:
    print "NO"
##n,k=ar()
##f=sorted(ar())
##t=0
##i=n-1
##while True:
##    #print i
##    t=t+((f[i]-1)*2)
##    if i-k>=0:
##        i-=k
##    else:
##        i-=k
##        break
###print i
####if i==0:
####    t=t+((f[i]-1)*2)
##print t
    

##n=int(raw_input())
##a=4
##b=9
##if n%2==0:
##    print n-a,a
##else:
##    print n-b,b