Problem:
Score : 400 points

Problem Statement

You are given an undirected connected weighted graph with N vertices and M edges that contains neither self-loops nor double edges.

The i-th (1≤i≤M) edge connects vertex a_i and vertex b_i with a distance of c_i.

Here, a self-loop is an edge where a_i = b_i (1≤i≤M), and double edges are two edges where (a_i,b_i)=(a_j,b_j) or (a_i,b_i)=(b_j,a_j) (1≤i<j≤M).

A connected graph is a graph where there is a path between every pair of different vertices.

Find the number of the edges that are not contained in any shortest path between any pair of different vertices.

Constraints

2≤N≤100

N-1≤M≤min(N(N-1)/2,1000)

1≤a_i,b_i≤N

1≤c_i≤1000

c_i is an integer.

The given graph contains neither self-loops nor double edges.

The given graph is connected.

Input

The input is given from Standard Input in the following format:

N M
a_1 b_1 c_1
a_2 b_2 c_2
:
a_M b_M c_M

Output

Print the number of the edges in the graph that are not contained in any shortest path between any pair of different vertices.

Sample Input 1

3 3
1 2 1
1 3 1
2 3 3

Sample Output 1

1

In the given graph, the shortest paths between all pairs of different vertices are as follows:

The shortest path from vertex 1 to vertex 2 is: vertex 1 → vertex 2, with the length of 1.

The shortest path from vertex 1 to vertex 3 is: vertex 1 → vertex 3, with the length of 1.

The shortest path from vertex 2 to vertex 1 is: vertex 2 → vertex 1, with the length of 1.

The shortest path from vertex 2 to vertex 3 is: vertex 2 → vertex 1 → vertex 3, with the length of 2.

The shortest path from vertex 3 to vertex 1 is: vertex 3 → vertex 1, with the length of 1.

The shortest path from vertex 3 to vertex 2 is: vertex 3 → vertex 1 → vertex 2, with the length of 2.

Thus, the only edge that is not contained in any shortest path, is the edge of length 3 connecting vertex 2 and vertex 3, hence the output should be 1.

Sample Input 2

3 2
1 2 1
2 3 1

Sample Output 2

0

Every edge is contained in some shortest path between some pair of different vertices.

Solution in Python (3.4.3) (judge verdict: Wrong Answer)

import copy

class MakeGraph:
    
    # 隣接行列もしくは隣接リストの初期化
    # node_num：ノード数
    # isMat：隣接行列(True) / 隣接リスト(False)
    def __init__(self, node_num:int, isMat:bool) -> None:
        self.n = node_num
        self.isMat = isMat

        # 隣接行列の初期化
        if self.isMat:
            INF = float("inf")
            self.graph = [[INF for _ in range(self.n)] for _ in range(self.n)]
    
        # 隣接リストの初期化
        else:
            self.graph = [[] for _ in range(self.n)]

    # エッジの追加
    # frm, to：起点ノード, 終点ノード
    # dist：frm→toへのコスト
    def add_undr_edge_mat(self, frm:int, to:int, dist:int) -> None:
        self.graph[frm][to] = dist
        self.graph[to][frm] = dist
        
    def add_drct_edge_mat(self, frm:int, to:int, dist:int) -> None:
        self.graph[frm][to] = dist
        
    def add_undr_edge_lst(self, frm:int, to:int, dist:int) -> None:
        self.graph[frm].append((dist, to))
        self.graph[to].append((dist, frm))
        
    def add_drct_edge_lst(self, frm:int, to:int, dist:int) -> None:
        self.graph[frm].append((dist, to))

                
    # 完成したグラフを取り出す
    def get_graph(self) -> list:
        return self.graph


N,M = map(int, input().split())
ABC = []
for i in range(M):
    ABC.append(tuple(map(int, input().split())))
    
graphMaker = MakeGraph(N,True)
add_edge = graphMaker.add_undr_edge_mat
for a,b,c in ABC:
    add_edge(a-1, b-1, c)
    
graph = graphMaker.get_graph()
for i in range(N):
    graph[i][i] = 0

dist = copy.deepcopy(graph)


# WF        
for i in range(N):
    for j in range(N):
        for k in range(N):
            if dist[i][j] > dist[i][k]+dist[k][j]:
                dist[i][j] = dist[i][k]+dist[k][j]
 
# 辺集合
edge_list = []
for a,b,c in ABC:
    edge_list.append([a-1,b-1])   

ans = 0
for i,j in edge_list:
    flag = 1
    for frm in range(N):
        for to in range(N):
            if frm == to:
                continue

            if dist[frm][i]+graph[i][j]+dist[j][to] == dist[frm][to]:
                flag = 0
                    
    ans += flag
  
print(ans)