human
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,...

#pri
assistant
#Hereever handle is chosen, I must be the first, then tourist and Petr must be the last.
nt cmp(2, 2)

class csj:
    def __init__(self, s, f):
        self.s = s
        self.f = f
        #self.rank = rank

n = int(raw_input())
csjs = []
for i in range(n):
    st = raw_input().split()
    #print st
    csjs.append(csj(st[0], st[1]))
rank = map(int, raw_input().split())
ok = 1
pre = ''
for i in range(n):
    #print pre
    ra = rank[i] - 1
    if(cmp(csjs[ra].s, csjs[ra].f) > 0):
        csjs[ra].s, csjs[ra].f = csjs[ra].f, csjs[ra].s
    if(pre == '' or cmp(pre, csjs[ra].s) < 0):
        pre = csjs[ra].s
    elif (cmp(pre, csjs[ra].f) < 0):
        pre = csjs[ra].f
    else:
        ok = 0
        break    
#al = s + f
#al.sort()

#print al

#csjs = []
#for i in range(n):
    #csjs.append(csj(s[i], f[i], rank[i]))

#for i in range(n):
    #print csjs[i].s, csjs[i].f, csjs[i].rank

#csjs.sort(lambda x,y: cmp(x.rank, y.rank))

#for i in range(n):
    #print csjs[i].s, csjs[i].f, csjs[i].rank
#ok = 1
#pre = ''
#for i in range(n):
    #if(i == 0):
        #if(cmp(csjs[i].s, csjs[i].f) > 0):
            #csjs[i].s, csjs[i].f = csjs[i].f, csjs[i].s
    #if(pre == '' or cmp(pre, csjs[i].s) < 0):
        #pre = csjs[i].s
    #elif (cmp(pre, csjs[i].f) < 0):
        #pre = csjs[i].f
    #else:
        #ok = 0
        #break
if (ok == 0):
    print 'NO'
else:
    print 'YES'