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 (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
 
int main(int argc, const char * argv[]) {
    // insert code here...
    int n = 0, i, j, cnt=0, save=0, connect=0;//connect=0初回  ３連続でない 7連続である
    int s[6][2];
   
    scanf("%d",&n);
    
    for (i = 0; i < 6; i++) {
        for (j = 0; j < 2; j++) {
            scanf("%d", &s[i][j]);
        }
    }
    
    
//    printf("%d", n);
//    printf("\n");
//    for (i = 0; i < n; i++) {
//        for (j = 0; j < 2; j++) {
//            printf("%d ", s[i][j]);
//        }
//        printf("\n");
//    }
    for(i = 0; i < n; i++){
        if(s[i][0]==s[i][1]){
            if(connect==0||connect==7){
                cnt++;
            }else{
                cnt=1;
            }
            connect=7;
        }else{
           
            if(save<cnt){
                save = cnt;
            }
        
            connect=3;
        }
    }
    if(save>cnt){
        cnt=save;
    }
  
    if(cnt>=3){
        printf("Yes\n");
    }else{
        printf("No\n");
    }
    return 0;
}