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>
#include<ext/rope>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#define mst(a,b) memset(a,b,sizeof(a))
#define ALL(x) x.begin(),x.end()
#define pii pair<int,int>
#define mp make_pair
#define lowbit(x) (x&(-x))
#define X first
#define Y second
using namespace __gnu_pbds;
using namespace __gnu_cxx;
using namespace std;
typedef long long ll;
typedef long long LL;
typedef unsigned long long ull;
typedef unsigned int ui;
typedef pair<ll, ll> pll;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const long double eps = 1e-8;
const long double pi = acos(-1.0);
const int N = 1e6 + 5;
const int mod = 1e9 + 21;
const int maxn = 1e7 + 5;
int e[4][4];
int main() {
#ifdef local
    freopen("input", "r", stdin);
//    freopen("out.txt","w",stdout);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    for(int i = 1; i <= 3; i++){
        int u,v;
        cin>>u>>v;
        e[u][v] = 1;
        e[v][u] = 1;
    }
    int a[4] = {0,1,2,3};
    do{
        int b = 1;
        for(int i = 2; i <= 4; i++){
            if(!e[a[i]][a[i - 1]])b = 0;
        }
        if(b){
            cout<<"YES"<<endl;
            return 0;
        }
    }while(next_permutation(a + 1,a + 4));
    cout<<"NO"<<endl;
}