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)

/*  CREATED BY
    STREAM_CIPHER
    19-sep-2020
*/
#include<bits/stdc++.h>
using namespace std;
#define int long long int
#define double long double
#define debug1(a) cout<<#a<<"="<<(a)<<"\n"
#define debug2(a,b) cout<<#a<<"="<<(a)<<","<<#b<<"="<<(b)<<"\n"
#define debug3(a,b,c) cout<<#a<<"="<<(a)<<","<<#b<<"="<<(b)<<","<<#c<<"="<<(c)<<"\n"
#define fix_precision(n) cout<<fixed<<setprecision(n)
#define all(a) a.begin(),a.end()
const double pi=acos(-1.0);
int inf=0x3f3f3f3f3f3f3f3f;
const int mod=1e9+7;
const int mx=5*1000000;//5*64M bit ->5*8M byte ->40MB size for long long int (64 bit)
int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    #ifndef ONLINE_JUDGE
        freopen("input.txt","r",stdin);
        // freopen("output.txt","w",stdout);
    #endif
        int n;
        cin>>n;
        vector<pair<int,int>>a(n);
        vector<int>aa(n);
        for(int i=0;i<n;i++)
        {
            cin>>a[i].first>>a[i].second;
            if(a[i].first==a[i].second)
                aa[i]=1;
            else
                aa[i]=0;
        }
        for(int i=0;i<n-2;i++)
        {
            if(aa[i]==aa[i+1]&&aa[i+1]==aa[i+2]&&aa[i]==1)
            {
                cout<<"Yes\n";
                return 0;
            }
        }
        cout<<"No\n";
}