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: Accepted)

N,M=list(map(int,input().split()))
e_list = [[] for i in range(N)]
e_list2 = []
for i in range(M):
    a,b,d = list(map(int,input().split()))
    a=a-1
    b=b-1
    e_list[a].append([b,d])
    e_list[b].append([a,d])
    e_list2.append([a,b,d])

dist_list = []
import heapq
for i in range(N):
    vi = i
    min_d_list = [10**27 for i in range(N)]
    min_d_list[vi] = 0
    prev_list = [-1 for i in range(N)]
    q = []
    for i in range(N):
        heapq.heappush(q,[min_d_list[i],i])

    while len(q)>0:
        d,v = heapq.heappop(q)
        for e in e_list[v]:
            v1,d1=e
            if min_d_list[v1]>d+d1:
                min_d_list[v1]=d+d1
                prev_list[v1]=v
                heapq.heappush(q,[d+d1,v1])
    dist_list.append(min_d_list)
count=len(e_list2)
for e in e_list2:
    for i in range(N):
        if dist_list[i][e[0]]+e[2]==dist_list[i][e[1]]:
            count-=1
            break

print(count)