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<iostream>
#include<algorithm>
#include<ctime>
#include<vector>
#include<string>
#include<cmath>
#include<map>
#include<iomanip>
#include<numeric>
#include<queue>
#include<deque>
#include<cfloat>
#include<functional>
#include<tuple>
#include<math.h>
#include<bitset>
#include<stack>
#include<set>
#include<random>
#include<stdlib.h>
#define rip(i,n) for(int i=0;i<n;i++)
#define Rip(i,n) for(int i=1;i<=n;i++)
#define RIP(i,a,b) for(int i=a;i<b;i++)
#define all(V) V.begin(),V.end()
#define sec  setprecision;
#define _CRT_SECURE_NO_WARNINGS
#pragma target("avx")
#pragma optimize("O3")
#pragma optimize("unroll-loops");
constexpr double eps = 1e-9;
constexpr double pi = 3.141592653589793238462643383279;
using namespace std;
using ll = long long;
using ld = long double;
using Pa = pair<int, int>;
ll mod = 1000000007;
ll MOD = 998244353;
ll gcd(ll a, ll b)
{
	if (a % b == 0)	return(b);
	else return(gcd(b, a % b));
}
ll lcm(ll a, ll b)
{
	return a * b / gcd(a, b);
}
ll pow(ll x, ll n) {
	x %= mod;
	if (n == 0)
		return 1;
	if (n % 2 == 0)
		return (pow((x * x) % mod, n / 2) % mod);
	else
		return (x * pow(x, n - 1)) % mod;
}
int main() {
	int n; cin >> n;
	vector<bool>r(n,false);
	rip(i, n) {
		int a, b; cin >> a >> b;
		if (a == b)r[i] = true;
	}
	rip(i, n - 2) {
		if (r[i] && r[i + 1] && r[i + 2]) {
			cout << "Yes" << endl;
			return 0;
		}
	}
	cout << "No" << endl;
}