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

#define rep(i,x,y) for(ll i=(x);i<(y);i++)
#define rrep(i,x,y) for(ll i=(ll)(y)-1;i>=(x);i--)
#define all(x) (x).begin(),(x).end()
#define itrout(x) for(int i=0;i<x.size();i++) {cout << x[i] << (i==x.size()-1 ? "\n" : " ");}
#ifdef LOCAL
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl
#define debugbit(x, n) cerr << #x << " = " << bitset<n>(x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl
#define itrdebug(x) cerr << #x << " "; for (auto & el : (x)) {cerr << (el) << " ";} cerr << endl
#define dassert(...) assert(__VA_ARGS__)
#else
#define debug(x)
#define debugbit(x, n)
#define itrdebug(x)
#define dassert(...)
#endif
//#define int long long
//using mint = atcoder::modint;

typedef long long ll;
const ll MOD = 1e9 + 7;
const long double EPS = 1e-8;

void solve(long long A, long long B){
    ll X = (A + B) / 2, Y = (A - B) / 2;
    cout << X << " " << Y << endl;
}

signed main(){
    // ios_base::sync_with_stdio(false);
    // cin.tie(NULL);
    long long A;
    scanf("%lld",&A);
    long long B;
    scanf("%lld",&B);
    solve(A, B);
    return 0;
}

//#include <atcoder/all>
#include <bits/stdc++.h>
//using namespace atcoder;
using namespace std;
typedef long long ll;

#define MOD (long long)(1e9+7)
#define INF (1LL<<60)
#define rep(i,n) for(ll i = 0; i < (n); i++)
#define rep1(i,n) for(ll i = 1; i <= (n); i++)

template<class T> inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template<class T> inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}


// 最大公約数
ll gcd(ll a, ll b)
{
  if(b == 0) return a;
  return gcd(b, a % b);
}

// mod m におけるa の逆元
ll modinv(ll a, ll m) {
  ll b = m, u = 1, v = 0;
  while (b) {
    ll t = a / b;
    a -= t * b; swap(a, b);
    u -= t * v; swap(u, v);
  }
  u %= m;
  if (u < 0) u += m;
  return u;
}

// 素因数分解
vector<pair<ll, ll>> prim;
void pf(ll n)
{
  ll s = sqrt(n);
  ll r = 0;
  for(ll i = 2; i <= s; i++) {
    if((n % i) == 0) {
      r = 0;
      do {
	r++;
	n = n / i;
      } while((n % i) == 0);
      prim.push_back({i, r});
    }
  }
  if(n > s) {
    prim.push_back({n, 1});
  }
}

void solve()
{
  ll N,W; cin >> N>>W;
   bool ok = true;
  // ll N, K; cin >> N >> K;
  // ll x, y, z; cin >> x >> y >> z;
  // ll a, b, c; cin >> a >> b >> c;
  // string s; cin >> s;
  vector<ll> s(N),t(N),p(N); rep(i, N) cin >> s[i]>>t[i]>>p[i];
  vector<ll> im(200200,0);
  rep(i,N){
    im[s[i]]+=p[i];
    im[t[i]]-=p[i];
  }
  ll su=0;
  rep(i,200000){
    su+=im[i];
    if(su>W)ok=false;
  }
  
  
  cout << (ok?"Yes":"No") << endl;
}


int main(void)
{
  // ll t; cin >> t; rep(i, t)
  solve();
}
