problem

There are $ N $ islands numbered from $ 1 $ to $ N $.
Each island has $ N-1 $ bridges, allowing any $ 2 $ island to move to each other across several bridges.
Each bridge has durability, and the durability of the $ i $ th bridge given the input is $ w_i $.
There are $ 1 $ treasures on each island, and you can pick them up while you're on the island.

Yebi, who is currently on the island $ S $, wants to carry all the treasures to the museum on the island $ E $.
Since yebi has ✝ magical power ✝, every time he visits the island $ v $, the durability of all bridges coming out of $ v $ is reduced by $ T $.
When the durability of the bridge drops below $ 0 $, the bridge collapses and cannot be crossed after that.
Can yebi deliver all the treasures to the museum?
However, since yebi is powerful, he can carry as many treasures as he wants at the same time.



output

Output "Yes" if you can deliver all the treasures to the museum, otherwise output "No".
Also, output a line break at the end.

Example

Input

4 10 1 4
1 2 52
1 3 68
3 4 45


Output

Yes
#define NAAN_TIKE_PLATYPUS
#define _CRT_SECURE_NO_WARNINGS

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <cfloat>
#include <cstring>
#include <map>
#include <utility>
#include <set>
#include <iostream>
#include <memory>
#include <string>
#include <vector>
#include <list>
#include <algorithm>
#include <functional>
#include <sstream>
#include <complex>
#include <stack>
#include <queue>
#include <unordered_set>
#include <unordered_map>
#include <array>
#include <cassert>
#include <bitset>
#include <cstdint>
using namespace std;
using LL = long long;


struct edge {
	int to;
	long long cost;
	int isE;
	edge(int to_, long long cost_, int isE_) :
		to(to_), cost(cost_), isE(isE_) {}
};
const long long INF = 123456789012345;
using P = pair<long long, int>;
void dijkstra(vector<edge>graph_[], long long dist[], int n_, int s_) {
	priority_queue<P, vector<P>, greater<P>>que;
	que.push({ 0,s_ });
	fill(dist, dist + n_, INF);
	dist[s_] = 0ll;
	while (!que.empty()) {
		auto q = que.top();
		que.pop();
		long long thd = q.first;
		int thv = q.second;
		if (thd > dist[thv])
			continue;
		for (auto ed : graph_[thv]) {
			if (dist[ed.to] > ed.cost + thd) {
				dist[ed.to] = ed.cost + thd;
				que.push({ dist[ed.to],ed.to });
			}
		}
	}
}

LL N, T, S, E;
vector<edge>G[123456];
bool mark[123456];
LL cnt[123456];
bool done = false;
bool suc = false;
bool used[123456];
void dfs(int x) {
	//cout << x << endl;
	used[x] = true;
	if (done)return;
	++cnt[x];
	for (auto& p : G[x]) {
		int& v = p.to;
		if (used[v])continue;
		if (p.cost <= (cnt[x] + cnt[p.to]) * T) {
			//渡れない
			done = true;
			return;
		}
		dfs(v);
		if (done)return;
		if (p.cost <= (cnt[x] + cnt[p.to]) * T) {
			//渡れない
			done = true;
			return;
		}
		if (done)return;
		++cnt[x];
	}
	if (x == E) {
		done = true;
		suc = true;
	}
}

LL dS[123456], dE[123456];

int main() {
	cin >> N >> T >> S >> E;
	--S; --E;
	for (int i = 1; i < N; ++i) {
		LL a, b, w;
		cin >> a >> b >> w;
		--a; --b;
		G[a].push_back({ int(b),w,0 });
		G[b].push_back({ int(a),w,0 });
	}
	dijkstra(G, dS, N, S);
	dijkstra(G, dE, N, E);
	LL SE = dS[E];
	for (int i = 0; i < N; ++i) {
		if (dS[i] + dE[i] == SE) {
			mark[i] = true;
		}
	}
	for (int i = 0; i < N; ++i) {
		for (auto& p : G[i]) {
			p.isE = (mark[p.to] && mark[i]);
		}
	}
	for (int x = 0; x < N; ++x) {
		sort(G[x].begin(), G[x].end(), [&](const edge& e1, const edge& e2) {if (e1.isE == e2.isE)return e1.cost > e2.cost; return e1.isE < e2.isE; });
	}
	cnt[S] = -1;
	dfs(S);
	if (suc) {
		cout << "Yes" << endl;
	}
	else {
		cout << "No" << endl;
	}
	return 0;
}
