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)

#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
#include <bits/stdc++.h> 
using namespace std;
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define rep(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
#define INF 1e18L+5
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define make_unique(x) sort(all((x))); (x).resize(unique(all((x))) - (x).begin())
typedef vector<int> vi;
typedef pair<int,int>ii;
typedef vector<vi> vvi;
typedef vector<ii> vii;
typedef long long ll;
typedef vector<long long > vl;
template<typename T>
void show(vector<T> v){
    bool ok=true;
for(auto it:v){

    
   cout<<it<<" ";
    //cout<<it;
}
cout<<"\n";
}
const int N=101;
int A[N],B[N];
 int main() { ios_base::sync_with_stdio(false); cin.tie(0);
 int n;cin>>n;
 rep(i,0,n)cin>>A[i]>>B[i];bool ok=false;
 for(int i=0;i<n;i++){
     if(A[i]==B[i]){
         int compt=1;
         rep(j,1,3){
             if(i+j>=n)continue;
             if(A[i+j]==B[i+j])++compt;
         }
         if(compt==3){
             cout<<"Yes";
             return 0;
         }
     }
 }
 puts("No");
return 0;
}