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 (2.7.6) (judge verdict: Accepted)

from heapq import heappush, heappop
n, m = map(int, raw_input().split())
G = [[] for i in xrange(n)]
E = {}
for i in xrange(m):
    a, b, c = map(int, raw_input().split())
    G[a-1].append((b-1, c))
    G[b-1].append((a-1, c))
    E[a-1, b-1] = E[b-1, a-1] = [c, 0]
def dijkstra(s):
    dist = [10**18]*n
    prev = [-1]*n
    que = [(0, s)]
    dist[s] = 0
    dic = set()
    while que:
        co, v = heappop(que)
        if dist[v] < co:
            continue
        for t, cost in G[v]:
            if co + cost < dist[t]:
                dist[t] = co + cost
                heappush(que, (co+cost, t))
                prev[t] = v
    for i in xrange(n):
        if prev[i] != -1:
            dic.add((i, prev[i]))
    return dic

ans = m
for i in xrange(n):
    dic = dijkstra(i)
    for k in dic:
        e = E[k]
        if e[1] == 0:
            ans -= 1
            e[1] = 1
print ans