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)

//B - Go to Jail
//Hiia
#include <stdio.h>
#include <stdlib.h>//rannsuu, zettaichi(absolute)
#include <time.h>
//srand((unsigned int) time(0));
#include <math.h>//koudona keisann
#include <string.h>//hairetsu
#define max(p,q)((p)>(q)?(p):(q))
#define min(p,q)((p)<(q)?(p):(q))
typedef long long ll;
//#define N 998244353
#define M 1000000007
#define pi 3.1415926535897932384626433832

int gcd(int m,int n){if((0==m)||(0==n))return 0;while(m!=n)if(m>n)m=m-n;else n=n-m;return m;}
int lcm(int m,int n){if((0==m)||(0==n))return 0;return(m*n/gcd(m, n));}
int fact(int x){if(x==1) return(x);else return(x*fact(x-1));}//Kaijou x!
int ruijou(int base,int exponent){int i,answer=1;i=exponent;while(i!=0){answer=answer*base;--i;}return answer;}//ruijou base^exponent

long long int llgcd(long long int m,long long int n){if((0==m)||(0==n))return 0;while(m!=n)if(m>n)m=m-n;else n=n-m;return m;}
long long int lllcm(long long int m,long long int n){if((0==m)||(0==n))return 0;return(m*n/llgcd(m, n));}
long long int llfact(long long int x){if(x==1) return(x);else return(x*llfact(x-1));}//Kaijou
int llruijou(long long int base,long long int exponent){long long int i,answer=1;i=exponent;while(i!=0){answer=answer*base;--i;}return answer;}//ruijou base^exponent

int main()
{
  ll n,a[10010], b[10010],ans=0;
  scanf("%lld", &n);
  for(ll i = 1; i <= n; i++) {
    scanf("%lld %lld", &a[i], &b[i]);
  }

  for(ll i = 1; i <= n-2; i++) {
    if(a[i]==b[i] && a[i+1]==b[i+1] && a[i+2]==b[i+2]){
      ans++;
    }
  }

  if(ans>=1){
    printf("Yes.\n");
  }
  else{
    printf("No.\n");
  }

  return 0;
}