Problem:
Score : 700 points

Problem Statement

We have a graph with N vertices and M edges, and there are two people on the graph: Takahashi and Aoki.

The i-th edge connects Vertex U_i and Vertex V_i.
The time it takes to traverse this edge is D_i minutes, regardless of direction and who traverses the edge (Takahashi or Aoki).

Takahashi departs Vertex S and Aoki departs Vertex T at the same time. Takahashi travels to Vertex T and Aoki travels to Vertex S, both in the shortest time possible.
Find the number of the pairs of ways for Takahashi and Aoki to choose their shortest paths such that they never meet (at a vertex or on an edge) during the travel, modulo 10^9 + 7.

Constraints

1 \leq N \leq 100 000

1 \leq M \leq 200 000

1 \leq S, T \leq N

S \neq T

1 \leq U_i, V_i \leq N (1 \leq i \leq M)

1 \leq D_i \leq 10^9 (1 \leq i \leq M)

If i \neq j, then (U_i, V_i) \neq (U_j, V_j) and (U_i, V_i) \neq (V_j, U_j).

U_i \neq V_i (1 \leq i \leq M)

D_i are integers.

The given graph is connected.

Input

Input is given from Standard Input in the following format:

N M
S T
U_1 V_1 D_1
U_2 V_2 D_2
:
U_M V_M D_M

Output

Print the answer.

Sample Input 1

4 4
1 3
1 2 1
2 3 1
3 4 1
4 1 1

Sample Output 1

2

There are two ways to choose shortest paths that satisfies the condition:

Takahashi chooses the path 1 \rightarrow 2 \rightarrow 3, and Aoki chooses the path 3 \rightarrow 4 \rightarrow 1.

Takahashi chooses the path 1 \rightarrow 4 \rightarrow 3, and Aoki chooses the path 3 \rightarrow 2 \rightarrow 1.

Sample Input 2

3 3
1 3
1 2 1
2 3 1
3 1 2

Sample Output 2

2

Sample Input 3

3 3
1 3
1 2 1
2 3 1
3 1 2

Sample Output 3

2

Sample Input 4

8 13
4 2
7 3 9
6 2 3
1 6 4
7 6 9
3 8 9
1 2 2
2 8 12
8 6 9
2 5 5
4 2 18
5 3 7
5 1 515371567
4 8 6

Sample Output 4

6

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)


import java.util.*;
import java.io.*;

class Edge {
	int to, cost;
	public Edge(int to, int cost) {
		super();
		this.to = to;
		this.cost = cost;
	}
}

class Vertex{
	int id;
	long dist;
	public Vertex(int id, long dist) {
		super();
		this.id = id;
		this.dist = dist;
	}
}

public class Main {
	
	static List<Edge>[] graph; 
	static int mod = (int)1e9+7;
	
	public static void main(String[] args) {
		FastScanner fs = new FastScanner(System.in);
		int n = fs.nextInt(), m = fs.nextInt();
		int s = fs.nextInt()-1, t = fs.nextInt()-1;
		graph = new List[n];
		for (int i=0;i<n;i++) graph[i] = new ArrayList<>();
		for (int i=0;i<m;i++) {
			int u = fs.nextInt()-1 , v = fs.nextInt()-1, d = fs.nextInt();
			graph[u].add(new Edge(v, d)); 
			graph[v].add(new Edge(u, d)); 
		}
		long distS[] = new long[n]; //各ノードまでの距離
		long distT[] = new long[n];
		long numS[] = new long[n]; //ノードに最短で辿り着く時の経路の数
		long numT[] = new long[n];

		dijkstra(s, distS, numS);
		dijkstra(t, distT, numT);
		
		//SからT, TからSの全通りの行き方
		long total = distS[t];
		long all = numS[t] * numT[s] % mod;
		//点で出会う時　最短経路状の頂点かつ、二人からの距離が等しい
		long vNum = 0;
		for(int i=0;i<n;i++) {
			if(distS[i] == distT[i] && distS[i]*2 == total) {
				long tmp = numS[i] * numT[i] % mod;
				tmp = tmp * tmp % mod;
				vNum = ( vNum + tmp ) % mod;
			}
		}
		//辺で出会う時
		long eNum = 0;
		for(int i=0;i<n;i++) {
			for(Edge e : graph[i]) {
				int from = i, to = e.to;
				if(distS[from] + distT[from] == total && distT[to] * 2 < total && distT[from] * 2 > total) {
					long tmp = numS[from] * numT[to] % mod;
					tmp = tmp * tmp % mod;
					eNum = ( eNum + tmp ) % mod;
				}
			}
		}
		long ans = (all - ((vNum + eNum)%mod + mod)%mod)%mod; 
		System.out.println(ans);
	}
	static void dijkstra(int v, long dist[], long num[]) {
		long INF = Long.MAX_VALUE / 3;
		Arrays.fill(dist, INF);
		dist[v] = 0;
		num[v] = 1;
		PriorityQueue<Vertex> pq = new PriorityQueue<>((v1, v2)->v1.dist < v2.dist ? -1:1);
		pq.add(new Vertex(v, 0));
		while(!pq.isEmpty()) {
			Vertex now = pq.poll();
			if(now.dist > dist[now.id])continue;
			for(Edge e : graph[now.id]) {
				if(dist[e.to] > dist[now.id] + e.cost) {
					dist[e.to] = dist[now.id] + e.cost; 
					pq.add(new Vertex(e.to, dist[e.to]));
					num[e.to] = num[now.id];
				}
				else if( dist[e.to] == dist[now.id] + e.cost ){
					num[e.to] += num[now.id];					
				}
			}
		}
		
	}
}

//高速なScanner
class FastScanner {
	private BufferedReader reader = null;
	private StringTokenizer tokenizer = null;

	public FastScanner(InputStream in) {
		reader = new BufferedReader(new InputStreamReader(in));
		tokenizer = null;
	}

	public String next() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(reader.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}

	public String nextLine() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken("\n");
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}

	public int[] nextIntArray(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}

	public long[] nextLongArray(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLong();
		return a;
	}
}