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("O3")
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
using namespace ::std;
// clang-format off
template<typename T> T nxt() {T res; cin >> res; return res;}
#define tests() for (int _ = nxt<int>(); _ > 0; _--)
#define nxt1(t,n) t (n); cin >> n
#define nxt2(t,n,m) t (n),(m); cin >> n >> m
#define nxt3(t,n,m,k) t (n),(m),(k); cin >> n >> m >> k
#define rep(i, A) for (int i = 0 ; i < A ; i++)
#define ll long long
typedef pair<int,int> pii;typedef pair<ll, ll> pll;typedef vector<int> vi;typedef vector<vi> vvi;typedef vector<ll> vl;typedef vector<vl> vvl;
#define dbg(vari) cerr<<#vari<<" = "<<(vari)<<endl;
template<typename T,typename TT> ostream& operator<<(ostream &s,pair<T,TT> t) {return s<<"("<<t.first<<","<<t.second<<")";}
template<typename T> ostream& operator<<(ostream &s,vector<T> t){for(T ti : t)s<<ti<<" ";return s; }
#define all(x) (x).begin(), (x).end()
#define last(vec) vec[vec.size()-1]
#define has(c,x) ((c).find(x) != (c).end())
// clang-format on

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int N = nxt<int>();
    int dublet = 0;
    bool res = false;
    rep(i, N) {
        nxt2(int, D1, D2);
        if (D1 == D2) {
            dublet++;
            if (dublet == 3)
                res = true;
        } else {
            dublet = 0;
        }
    }

    cout << (res ? "Yes" : "No") << endl;
    return 0;
}