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 = map(int, input().split())
Map = [[int(1e18) for _ in range(N)] for __ in range(N)]
Path = [[[] for _ in range(N)] for __ in range(N)]
D = {}
for i in range(M):
  a, b, c = map(int, input().split())
  Map[a-1][b-1] = c
  Map[b-1][a-1] = c
  Path[a-1][b-1] = [a-1, b-1]
  Path[b-1][a-1] = [b-1, a-1]
  D[(a-1)*N+b-1] = 0
  D[(b-1)*N+a-1] = 0
for i in range(N):
  Map[i][i] = 0

for k in range(N):
  for i in range(N):
    for j in range(N):
      if Map[i][j] > Map[i][k] + Map[k][j]:
        Map[i][j] = Map[i][k] + Map[k][j]
        Path[i][j] = Path[i][k][:] + Path[k][j][1:]
        
for i in range(N):
  for j in range(N):
    X = Path[i][j][:]
    n = len(X)
    for k in range(n-1):
      l = X[k] * N + X[k+1]
      r = X[k+1] * N + X[k]
      if l in D:
        D[l] = 1
      if r in D:
        D[r] = 1
SUM41 = 0
for i in D:
  if D[i] == 0:
    SUM41 += 1
print(SUM41 // 2)