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)

#ifdef __GNUC__
#pragma GCC diagnostic ignored "-Wunused-result"
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#endif

#include <algorithm>
#include <cmath>
#include <cstring>
#include <iostream>
#include <sstream>
#include <numeric>
#include <map>
#include <set>
#include <queue>
#include <vector>

using namespace std;

typedef long long LL;
typedef long double LD;
typedef pair<LL, LL> II;
typedef pair<LL, II> III;
typedef priority_queue<III, vector<III>, greater<>> Queue;
static const LL INF = 1LL << 60;

LL gcd(LL a, LL b) { return b ? gcd(b, a % b) : a; }
LL lcm(LL a, LL b) { return (a / gcd(a, b)) * b; }

#if 0
	auto possible = [&](LL t) {

		return false;
	};
	LL left = 1, right = 1000000000;
	while (right - left > 1) {
		LL mid = (left + right) / 2;
		if (possible(mid)) {
			left = mid;
		} else {
			right = mid;
		}
	}
#endif

void solve(long long N, std::vector<std::vector<long long>> &D) {
	bool ans = false;
	int cnt = 0;
	for (LL i = 0; i < N; ++i) {
		if (D[i][0] == D[i][1]) {
			++cnt;
			if (cnt == 3) ans = true;
		} else {
			cnt = 0;
		}
	}
	cout << (ans ? "Yes" : "No") << endl;
}

int main() {
    long long N;
	std::cin >> N;
	std::vector<std::vector<long long>> D(N, std::vector<long long>(2));
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < 2; j++) {
			std::cin >> D[i][j]; // D[i][j]--;
		}
	}
	solve(N, D);
	return 0;
}