Problem:
Score : 200 points

Problem Statement

Tak performed the following action N times: rolling two dice.
The result of the i-th roll is D_{i,1} and D_{i,2}.

Check if doublets occurred at least three times in a row.
Specifically, check if there exists at lease one i such that D_{i,1}=D_{i,2}, D_{i+1,1}=D_{i+1,2} and D_{i+2,1}=D_{i+2,2} hold.

Constraints

3 \leq N \leq 100

1\leq D_{i,j} \leq 6

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
D_{1,1} D_{1,2}
\vdots
D_{N,1} D_{N,2}

Output

Print Yes if doublets occurred at least three times in a row. Print No otherwise.

Sample Input 1

5
1 2
6 6
4 4
3 3
3 2

Sample Output 1

Yes

From the second roll to the fourth roll, three doublets occurred in a row.

Sample Input 2

5
1 1
2 2
3 4
5 5
6 6

Sample Output 2

No

Sample Input 3

6
1 1
2 2
3 3
4 4
5 5
6 6

Sample Output 3

Yes

Solution in Kotlin (1.3.71) (judge verdict: Accepted)

import java.util.*

fun main() {
    val sc = Scanner(System.`in`)
    val n = sc.nextInt()
    val ds = (1..n*2).map { sc.nextInt() }

    println(calc(ds))
}

fun calc(ds: List<Int>): String {
    var cnt = 0
    var maxCnt = 0
    for(i in ds.indices step 2) {
        if(ds[i] == ds[i+1]) {
            cnt++
        }
        else {
            if(cnt > maxCnt) maxCnt = cnt
            cnt = 0
        }
    }
    if(cnt > maxCnt) maxCnt = cnt
    return if(maxCnt >= 3) "Yes" else "No"
}