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 heapq


def dist(N, a, b, w_mat):
    d_mat = [[100] * (N + 1) for _ in range(N + 1)]
    for i in range(N + 1):
        d_mat[i][i] = 0
    q = [(0, a)]
    heapq.heapify(q)
    # print("\n".join([str(r) for r in w_mat]))
    s = set()
    while len(q):
        # print(q)
        w, i = heapq.heappop(q)
        s.add(i)
        for ni, wi in [(ni, wi) for ni, wi in enumerate(w_mat[i])
                       if wi > 0 and ni not in s]:
            if d_mat[a][ni] > w + wi:
                d_mat[a][ni] = w + wi
                d_mat[ni][a] = w + wi
                heapq.heappush(q, (w + wi, ni))
    # print("\n".join([str(r) for r in d_mat]))
    # print("")
    return d_mat[a][b]


def solve(N, M, ABCs):
    # print(N, ",", M, ",", ABCs)
    d_mat = [[0] * (N + 1) for _ in range(N + 1)]

    w_mat = [[-1] * (N + 1) for _ in range(N + 1)]
    for i in range(N + 1):
        w_mat[i][i] = 0

    for a, b, c in ABCs:
        w_mat[a][b] = c
        w_mat[b][a] = c

    for s in range(1, N + 1):
        for e in range(s + 1, N + 1):
            d = dist(N, s, e, w_mat)
            d_mat[s][e] = d
            d_mat[e][s] = d

    ans = len(ABCs)
    for a, b, c in ABCs:
        if any(d_mat[s][e] == d_mat[s][a] + c + d_mat[b][e]
               for s in range(1, N + 1)
               for e in range(s + 1, N + 1)):
            ans -= 1

    #    print("\n".join([str(r) for r in d_mat]))
    return ans


if __name__ == "__main__":
    N, M = tuple(map(int, input().split(" ")))
    ABCs = [tuple(map(int, input().split(" "))) for _ in range(M)]
    print(solve(N, M, ABCs))