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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i, n) for(int i = 0; (i) < ((int)(n)); ++(i))
const ll INF = 1e10;
const ll MOD = 1e9 + 7;

int main() {
    int N;
    cin >> N;
    vector<vector<int>> D(N, vector<int>(2));
    rep(i, N) cin >> D[i][0] >> D[i][1];
    int Count = 0;
    rep(i, N) {
        if(D[i][0] == D[i][1])
            Count++;
        else
            Count = 0;

        if(Count >= 3) {
            cout << "Yes" << endl;
            return 0;
        }
    }

    cout << "No" << endl;
}