#abc247b. [abc247_b]Unique Nicknames

[abc247_b]Unique Nicknames

Problem Statement

There are NN people numbered Person 11, Person 22, dots\\dots, and Person NN. Person ii has a family name sis_i and a given name tit_i.

Consider giving a nickname to each of the NN people. Person ii's nickname aia_i should satisfy all the conditions below.

  • aia_i coincides with Person ii's family name or given name. In other words, ai=sia_i = s_i and/or ai=tia_i = t_i holds.
  • aia_i does not coincide with the family name and the given name of any other person. In other words, for all integer jj such that 1leqjleqN1 \\leq j \\leq N and ineqji \\neq j, it holds that aineqsja_i \\neq s_j and aineqtja_i \\neq t_j.

Is it possible to give nicknames to all the NN people? If it is possible, print Yes; otherwise, print No.

Constraints

  • 2leqNleq1002 \\leq N \\leq 100
  • NN is an integer.
  • sis_i and tit_i are strings of lengths between 11 and 1010 (inclusive) consisting of lowercase English alphabets.

Input

Input is given from Standard Input in the following format:

NN s1s_1 t1t_1 s2s_2 t2t_2 vdots\\vdots sNs_N tNt_N

Output

If it is possible to give nicknames to all the NN people, print Yes; otherwise, print No.


Sample Input 1

3
tanaka taro
tanaka jiro
suzuki hanako

Sample Output 1

Yes

The following assignment satisfies the conditions of nicknames described in the Problem Statement: a1=a_1 = taro, a2=a_2 = jiro, a3=a_3 = hanako. (a3a_3 may be suzuki, too.)
However, note that we cannot let a1=a_1 = tanaka, which violates the second condition of nicknames, since Person 22's family name s2s_2 is tanaka too.


Sample Input 2

3
aaa bbb
xxx aaa
bbb yyy

Sample Output 2

No

There is no way to give nicknames satisfying the conditions in the Problem Statement.


Sample Input 3

2
tanaka taro
tanaka taro

Sample Output 3

No

There may be a pair of people with the same family name and the same given name.


Sample Input 4

3
takahashi chokudai
aoki kensho
snu ke

Sample Output 4

Yes

We can let a1=a_1 = chokudai, a2=a_2 = kensho, and a3=a_3 = ke.