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)

#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<math.h>
#include<iomanip>
#include<set>
#include<numeric>
#include<cstring>
#include<cstdio>
#include<functional>

#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define REVERSE(v,n) reverse(v,v+n);
#define VREVERSE(v) reverse(v.begin(), v.end());
#define ll long long
#define pb(a) push_back(a)
#define INF 9999999999
#define m0(x) memset(x,0,sizeof(x))
#define fill(x,y) memset(x,y,sizeof(x))
#define print(x) cout<<x<<endl;
#define pe(x) cout<<x<<" ";
#define lb(v,n) lower_bound(v.begin(), v.end(), n);
//#define int long long


using namespace std;


int dy[4] = { 0,0,1,-1 };
int dx[4] = { 1,-1,0,0 };
int dxx[9] = { 0, 0,0,1,1,1,-1,-1,-1 };
int dyy[9] = { 0, 1,-1,0,1,-1,0,1,-1 };




ll gcd(ll x, ll y) {
	ll m = max(x, y), n = min(x, y);
	if (m%n == 0)return n;
	else return gcd(m%n, n);
}
ll lcm(ll x, ll y) {
	return x / gcd(x, y)*y;
}

long long power_mod(long long x, long long n, long long m) {
	if (n == 0)
		return 1;
	if (n % 2 == 0)
		return power_mod(x * x % m, n / 2, m);
	else
		return x * power_mod(x, n - 1, m) % m;
}

long long power(long long a, long long n) {//aのn乗を計算します。
	long long x = 1;
	while (n > 0) {//全てのbitが捨てられるまで。
		if (n & 1) {//1番右のbitが1のとき。
			x = x * a;
		}
		a = a * a;
		n >>= 1;//bit全体を右に1つシフトして一番右を捨てる。
	}
	return x;
}



long long nCr(long long n, long long r) {
	if (r > n / 2) r = n - r; // because C(n, r) == C(n, n - r)
	long long ans = 1;
	long long i;

	for (i = 1; i <= r; i++) {
		ans *= n - r + i;
		ans /= i;
	}

	return ans;
}



const int MAX = 510000;
const int MOD = 1000000007;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAX; i++) {
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}
}

// 二項係数計算
long long COM(int n, int k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

bool arr[100000100];
vector<ll>sosuu;
void Eratosthenes() {
	ll N = 10000010;
	int c = 0;
	for (int i = 0; i < N; i++) {
		arr[i] = 1;
	}
	for (ll i = 2; i < sqrt(N); i++) {
		if (arr[i]) {
			for (ll j = 0; i * (j + 2) < N; j++) {
				arr[i *(j + 2)] = 0;
			}
		}
	}

	for (ll i = 2; i < N; i++) {
		if (arr[i]) {
			sosuu.pb(i);
			//cout << sosuu[c] << " ";
			c++;
		}
	}
	//cout << endl;
	//cout << c << endl;
}

ll stoL(string s) {
	ll n = s.length();
	ll ans = 0;
	for (int i = 0; i < n; i++) {
		ans += power(10, n - i - 1)*(ll)(s[i] - '0');
	}
	return ans;
}



int main() {
	map<int,int>mp;
	REP(i, 3) {
		int a, b; cin >> a >> b;
		mp[a]++, mp[b]++;
	}
	for (auto x : mp) {
		if (x.second > 2) {
			print("NO");
			return 0;
		}
	}
	print("YES");
}