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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define __ <<" "<<
#define ___ <<" "
#define bash push_back
#define ALL(x) x.begin(),x.end()
// #define int long long
struct IoSetup {
    IoSetup() {
        cin.tie(0);
        ios::sync_with_stdio(false);
        cout << fixed << setprecision(10);
        cerr << fixed << setprecision(10);
    }
}IoSetup;

using Int = __int128_t;
using ll = long long;
using pii = pair<int, int>;
constexpr int INF = 0x3f3f3f3f;
constexpr long long LINF = 0x3f3f3f3f3f3f3f3fLL;
constexpr int SMOD = 1000000007;
constexpr int NMOD = 998244353;
constexpr int dx[]={1,0,-1,0,1,1,-1,-1};
constexpr int dy[]={0,-1,0,1,-1,1,-1,1};

inline bool inside(int x,int y,int w,int h){return (x>=0 && y>=0 && x<w && y<h);}
template<class T>bool chmax(T &a, const T&b){if(a<b)return(a=b,1);return 0;}
template<class T>bool chmin(T &a, const T&b){if(b<a)return(a=b,1);return 0;}


signed main() {
    
	int n;
	cin >> n;
	int d[1000][2]={};

	for(int i = 0; i < n; i++) {
		cin >> d[i][0] >> d[i][1];
	}

	for(int i = 0; i < n; i++) {
		if(d[i][0] == d[i][1] and d[i+1][0] == d[i+1][1] and d[i+2][0] == d[i+2][1]) {
			cout << "Yes\n";
			return 0;
		}
	}

	cout << "No\n";
    return 0;
}