#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ar array
const int N = 60005;
int n,k,w1[N],w2[N],s[N],visited[N];
multiset<ar<int,2>>g[N];
bitset<N * 20> dp;
ll d = 0;
void dfs(int u) {
visited[u] = 1;
if (!g[u].size()) return;
for (auto [v, cost] : g[u]) {
if (!visited[v]) {
d += cost;
dfs(v);
}
}
g[u].clear();
}
void solve() {
cin >> n >> k;
memset(visited,0,n * 2 + 5);
for (int i = 1; i <= n * 2;i++) cin >> w1[i] >> w2[i] >> s[i];
// making graph , let's nodes of this graph denotes places, and edges denotes people
for (int i = 1; i <= n * 2;i++) g[w1[i]].insert({w2[i] + n,s[i]}),g[w2[i] + n].insert({w1[i],-s[i]});
for (int i = 1; i <= n * 2;i++) {
if (g[i].empty()) { // if nobody wants this place
cout << "NO";
return;
}
}
queue<int> q;
for (int i = 1; i <= n * 2;i++) {
if (g[i].size() == 1) q.push(i); // if there's only one person who wants this place
}
d = 0;
while (!q.empty()) {
int u = q.front(); // u is place not person
q.pop();
if (g[u].size() == 0) {
cout << "NO";
return;
}
for (auto[v,cost] : g[u]) {
d += cost;
// cost will be opposite
g[v].erase(g[v].find({u,-cost}));
if (g[v].size() == 1) q.push(v);
}
g[u].clear();
}
vector<ll>vals;
if (d) vals.push_back(abs(d));
for (int i = 1; i <= n * 2;i++) {
// we know that every single node belong to exactly one componenent that is full of cycle, so we just go dfs , and remove this edges
if (!visited[i] && !g[i].empty()) {
d = 0;
g[i].erase(g[i].begin()); // this edge is redundant
dfs(i);
if (d) vals.push_back(abs(d));
}
}
// can we divide vals into two subset S and T , sum(S) - sum(T) <= k , sum(vals) - 2 * sum(S) <= k
ll sum = accumulate(vals.begin(),vals.end(),0ll);
dp[0] = true;
sort(vals.begin(),vals.end());
for (ll x : vals) dp |= (dp << x);
for (int i = 1; i <= sum;i++) {
if (dp[i] && abs(sum - 2 * i) <= k) {
cout << "YES";
return;
}
}
cout << "NO";
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
solve();
return 0;
}
