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: Accepted)

/*
    問題をよく読もう！
    論理的に考えよう！
    サンプルを確認しよう!
    絶対に諦めるな！
    工夫をしろ！
    配列は少し多めにとっておく

    Twitterは終わるまでログアウト！
    （間違えて解法をツイートしてはいけないから）

*/
//include
#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>
#include<stdio.h>
#include<cstring>
//namespace
using namespace std;
//繰り返し
#define REP(i, m, n) for(int i=(int)m; i<(int)n; ++i)
#define rep(i, n) REP(i, 0, n)
//イテレータ
#define all_range(C) begin(C), end(C)
//簡略化
typedef long long ll;
typedef pair<int,int> pint;
typedef pair<ll,int> pli;
typedef pair<string,int> pst;

const int inf = 1e9+7;
const ll longinf = 1LL<<60;
const ll mod = 1e9+7;
//最大最小
template<typename T1, typename T2> inline void chmin(T1 &a, T2 b) {if(a>b) a=b;}
template<typename T1, typename T2> inline void chmax(T1 &a, T2 b) {if(a<b) a=b;}

//関数

//最大公約数を求める
ll my_gcd(ll x, ll y) {
  ll r;

  while(y!=0) {
    r = x%y;
    x = y;
    y = r;
  }

  return x;
}

//最小公約数を求める
ll my_lcm(ll x, ll y) {

  return (x*y)/my_gcd(x,y);
}

//xのy乗を求める ※参考：10^18乗まで
ll my_pow(ll x, ll y) {
  ll sum = 1;

  while(y>0) {
    sum *= x;
  }

  return sum;
}

//main.cpp----------------------------


void solve();

ll a[3], b[3];
ll c[5] = {};

int main() {
  rep(i, 3) cin >> a[i] >> b[i];

  solve();

  return 0;
}

void solve() {
  bool flag = true;

  rep(i, 3) {
    c[a[i]] += 1;
    c[b[i]] += 1;
  }

  if(c[1]!=1) flag = false;
  if(c[2]!=2) flag = false;
  if(c[3]!=2) flag = false;
  if(c[4]!=1) flag = false;

  if(flag) cout << "YES" << endl;
  else cout << "NO" << endl;
}