Problem:
Score : 200 points

Problem Statement

There are four towns, numbered 1,2,3 and 4.
Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally.
No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads.

Determine if we can visit all the towns by traversing each of the roads exactly once.

Constraints

1 \leq a_i,b_i \leq 4(1\leq i\leq 3)

a_i and b_i are different. (1\leq i\leq 3)

No two roads connect the same pair of towns.

Any town can be reached from any other town using the roads.

Input

Input is given from Standard Input in the following format:

a_1 b_1
a_2 b_2
a_3 b_3

Output

If we can visit all the towns by traversing each of the roads exactly once, print YES; otherwise, print NO.

Sample Input 1

4 2
1 3
2 3

Sample Output 1

YES

We can visit all the towns in the order 1,3,2,4.

Sample Input 2

3 2
2 4
1 2

Sample Output 2

NO

Sample Input 3

2 1
3 2
4 3

Sample Output 3

YES

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

#define endl "\n"
#define all(x) x.begin(),x.end()
#define pb push_back
#define ff first 
#define ss second 
#define pii pair<int,int>
#define vi vector<int>
#define vc vector<char>
#define vii vector<pii>
#define vvi vector<vi>
#define fori(i,n) for(int i=1;i<=n;++i)
#define forn(i,n) for(int i=0;i<n;++i)
#define sz(x) int(x.size())
#define error(x) cout<<#x<<" "<<x<<endl
#define IOS  ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)

const int N=100*1000;
const int mod=1e9+7;

int min3(int a,int b,int c){return min(a,min(b,c));}
int max3(int a,int b,int c){return max(a,max(b,c));}
int min4(int a,int b,int c,int d){return min(min(a,b),min(c,d));}
int max4(int a,int b,int c,int d){return max(max(a,b),max(c,d));}


int32_t  main(){
	IOS;
  vii v;
  for(int i=0;i<3;++i){
    int u_,v_;cin>>u_>>v_;
    v.pb({u_,v_});
  }
  //ohk now brute force 
  for(int i=0;i<3;++i){
    for(int j=0;j<3;++j){
      for(int k=0;k<3;++k){
        if(i==j || j==k || k==i)continue;
        if(v[i].ss==v[j].ff && v[j].ss==v[k].ff){
          puts("YES");
          return 0;
        }
      }
    }
  }
  puts("NO");

  return 0;
}