#include <bits/stdc++.h>
#define fo(i, d, c) for (int i = d; i <= c; i++)
#define fod(i, c, d) for (int i = c; i >= d; i--)
#define maxn 600010
#define N 1010
#define fi first
#define se second
#define pb emplace_back
#define en cout << "\n";
#define bitcount(x) __builtin_popcountll(x)
#define pii pair<int, int>
#define vii vector<pii>
#define lb(x) x & -x
#define bit(i, j) ((i >> j) & 1)
#define offbit(i, j) (i ^ (1LL << j))
#define onbit(i, j) (i | (1LL << j))
#define vi vector<int>
#define all(x) x.begin(), x.end()
#define ss(x) (int)x.size()
template <typename T1, typename T2>
bool minimize(T1 &a, T2 b)
{
if (a > b)
{
a = b;
return true;
}
return false;
}
template <typename T1, typename T2>
bool maximize(T1 &a, T2 b)
{
if (a < b)
{
a = b;
return true;
}
return false;
}
using namespace std;
const int nsqrt = 450;
const int mod = 1e9 + 7;
void add(int &x, int k)
{
x += k;
x %= mod;
if (x < 0)
x += mod;
}
void del(int &x, int k)
{
x -= k;
x %= mod;
if (x < 0)
x += mod;
}
int cnt[maxn];
bool ok[maxn];
array<int, 3> edge[maxn];
int n, k;
set<int> ke[maxn];
vi things;
int total = 0;
int sum;
void dfs(int u, int val)
{
if(ke[u].empty())
{
if(sum != 0)
things.pb(abs(sum));
return;
}
auto [i,j,w] = edge[*ke[u].begin()];
int v = i + j - u;
sum += w * val;
ke[v].erase(ke[v].find(*ke[u].begin()));
ke[u].erase(ke[u].begin());
dfs(v,-val);
}
bitset<600010> f;
main()
{
#define name "TASK"
if (fopen(name ".inp", "r"))
{
freopen(name ".inp", "r", stdin);
freopen(name ".out", "w", stdout);
}
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cin >> n >> k;
fo(i, 1, 2 * n)
{
int u, v, w;
cin >> u >> v >> w;
edge[i] = {u, v + n, w};
ke[u].insert(i);
ke[v + n].insert(i);
ok[i] = 1;
}
queue<int> q;
fo(i,1,2 * n)
{
if(ss(ke[i]) == 0) return cout << "NO",0;
if(ss(ke[i]) == 1) q.push(i);
}
while(ss(q))
{
int u = q.front();
q.pop();
if(ss(ke[u]) == 0) return cout << "NO",0;
int it = *ke[u].begin();
int v = edge[it][0] + edge[it][1] - u;
if(v > n) sum -= edge[it][2];
else sum += edge[it][2];
ke[u].clear();
ke[v].erase(ke[v].find(it));
if(ss(ke[v]) == 1) q.push(v);
}
if(sum != 0)
things.pb(abs(sum));
fo(i, 1, 2 * n) if (ss(ke[i]))
{
sum = 0;
dfs(i, 1);
}
total = accumulate(all(things),0);
f[0] = 1;
for (int it : things)
f |= f << it;
fo(i, 0, 600000) if (f[i] and abs(total - 2 * i) <= k) return cout << "YES", 0;
cout << "NO";
}
