#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 60001;
int n, k, l[MX],r[MX],s[MX];
bitset<600001> b;
vi adj[MX];
int pre[MX], depth[MX];
bool visit[MX], cyc[MX];
pi oops;
int yes[2], no[2], sum;
vi v;
vpi V;
void dfs(int x) {
visit[x] = 1; v.pb(x);
for (auto a: adj[x]) if (a != pre[x]) {
int t = l[a]+r[a]+n-x;
if (visit[t]) {
if (depth[t] < depth[x]) oops = {x,a};
} else {
pre[t] = a; depth[t] = depth[x]+1;
dfs(t);
}
}
}
int getSum(int x) { return l[x]+r[x]+n; }
void genCyc() {
V = {oops};
int fin = getSum(oops.s)-oops.f;
do {
int nex = getSum(pre[oops.f])-oops.f;
V.pb({nex,pre[oops.f]});
oops.f = nex;
} while (oops.f != fin);
}
void solve(int x) {
v.clear(), V.clear();
F0R(i,2) yes[i] = no[i] = 0;
dfs(x);
int numEdge = 0;
for (int i: v) numEdge += sz(adj[i]);
if (numEdge != 2*sz(v)) {
cout << "NO";
exit(0);
}
genCyc();
for (auto i: V) cyc[i.f] = 1;
F0R(i,sz(V)) no[i&1] += s[V[i].s];
b = (b<<(no[0]))|(b<<(no[1]));
// cout << "HA " << x << " " << yes[0] << " " << no[0] << " " << yes[1] << " " << no[1] << "\n";
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> n >> k;
b[0] = 1;
FOR(i,1,2*n+1) {
cin >> l[i] >> r[i] >> s[i];
adj[l[i]].pb(i), adj[r[i]+n].pb(i);
sum += s[i];
}
FOR(i,1,2*n+1) if (!visit[i]) solve(i);
// cout << "HA\n";
F0R(i,sum+1) if (b[i] == 1 && abs(2*i-sum) <= k) {
cout << "YES";
exit(0);
}
cout << "NO";
}
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
