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<iostream>
#include<stdio.h>
#include<vector>//s1.erase(n)=文字列s1の、n文字目以降削除
#include<algorithm>//辞書順=next_permutation
#include<set>
#include<string>//char 小文字のほうが32大きい
#include<sstream>
#include<string.h>
#include<complex>
#include<time.h>
#include<random>
#include<cmath>
#include<math.h>
#include<queue>
#include<time.h>
#include<iomanip>
#include<locale>
#include<utility>//swap=数値交換
#define rt "\n"
#define rep(i,n) for(int i=0;i<n;i++)
#define rop(i,n) for(int i=1;i<=n;i++)
#define drep(i,n) for(int i=n-1;0<=i;i--)
#define drop(i,n) for(int i=n;0<i;i--)
#define yes(ans) if(ans)cout<<"yes"<<rt;else cout<<"no"<<rt;
#define Yes(ans) if(ans)cout<<"Yes"<<rt;else cout<<"No"<<rt;
#define YES(ans) if(ans)cout<<"YES"<<rt;else cout<<"NO"<<rt;
#define sec(a,b,ans) if(ans)cout<<a<<rt;else cout<<b<<rt;
#define vcin(s,n) rep(i,n)cin>>s[i];//vcin(配列名),(繰り返し回数)
#define sort(s) sort(s.begin(),s.end())//標準=昇順
#define reve(s) reverse(s.begin(),s.end())
#define asas int ans=0
#define llcncn llint cnt=0
#define llasas llint ans=0
#define str string
#define please return
#define AC 0
#define ic_a int a;cin>>a;
#define ic_b int b;cin>>b;
#define ic_s string s;cin>>s;
#define Rapid_pleaseAC_fast cin.tie(0);ios::sync_with_stdio(false)
#define Pi 3.1415926535897932384626
#define nine 1000000000

using namespace std;
typedef vector<int> vint;
typedef vector<string> vstr;
typedef vector<char> vchar;
typedef vector<double> vdou;
typedef long long int llint;
typedef pair<int, int> pint;
typedef pair<llint, llint> pllint;
typedef vector<llint> vllint;
typedef vector<pint> vpint;
typedef vector<pair<llint, llint>> vpllint;
typedef vector<vector<int>> vvint;
typedef vector<vector<char>> vvchar;
typedef vector<vector<llint>> vvllint;
typedef vector<vector<string>> vvstr;
typedef vector<vector<bool>> vvbool;
typedef vector<vector<pint>> vvpint;
typedef vector<bool> vbool;
typedef long double ld;
const ld eps = 1.0e-10; // 許容される誤差
long long GCD(long long a, long long b) {
	if (b == 0) return a;
	else return GCD(b, a % b);
}

long long LCM(long long a, long long b) {
	return a * b / GCD(a, b);
}

unsigned GetDigit(unsigned num) {
	return std::to_string(num).length();
}

int tow(int n) {//2のn乗
	if (n == 0)return 1;
	int x = tow(n / 2);
	x *= x;
	if (n % 2 == 1)x *= 2;
	return x;//@domino
}

int keta(int n) {
	int sum = 0;
	while (n != 0) {
		sum += n % 10; n /= 10;
	}
	return sum;//Nanashi,tRue,wai
}

/*

 (char)toupper(a[n])=文字列のn文字目を大文字で出力

  pow(a,b)=aのb乗

  */
ld mysqrtl(ld x)
{
	ld a = sqrtl((double)x); // 近似値
	do {
		a = (a + x / a) / 2.0L;
	} while (fabsl(x - a * a) > eps);
	return a;
}

//平方根を求める。sqrtより精度が高い。
//epsを変えて誤差を調整する。
int main(void) {
	Rapid_pleaseAC_fast;
	llint ans = 0, n = 0, m=1, count=0, sum = 0, ch = 0;
	llint  a, aa, aaa, rr = 0, gg = 0, bb = 0,b=0;
	llint bbb=0, bbbb=0;
	bool check = false;
	string s,out;
	vector<llint> r, g;
	char c;
	
	rep(i, 6) {
		cin >> a;
		if (a == 1)b++;
		if (a == 2)bb++;
		if (a == 3)bbb++;
		else bbbb++;
	}
	if (b <= 2 && bb <= 2 && bbb <= 2 && bbbb <= 2)cout << "YES" << endl;
	else cout <<"NO"<< endl;
	
	please AC;
}