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: Time Limit Exceeded)

# https://atcoder.jp/contests/abc051/tasks/abc051_d
# 想定解法
# 最短距離の隣接行列Dが与えられたときに、edges[a, b]が最短経路に含まれるかどうか考える
# 2点s,tについて、最短距離であれば、D[s,t]=D[s,i]+edges[i,j]+D[j,t]を満たすi,jの組み合わせは最短経路辺である
# 5000*1000の探索なので間に合う

from scipy.sparse import csr_matrix  # 自分で配列を作ってからcsrに入れよう(lilに打ち込んでいくのは非常に遅い)

# ダイクストラ法 (正の単一始点最短経路) (全点間最短経路)
from scipy.sparse.csgraph import dijkstra
import sys
sys.setrecursionlimit(1 << 25)
read = sys.stdin.readline
ra = range
enu = enumerate


def read_ints():
    return list(map(int, read().split()))


N, M = read_ints()
adj_mat = [[0] * N for _ in range(N)]
INF = 10 ** 9
from collections import defaultdict
from itertools import combinations
edges = defaultdict(lambda: INF)
for _ in range(M):
    a, b, c = read_ints()
    a -= 1
    b -= 1
    adj_mat[a][b] = c
    adj_mat[b][a] = c
    edges[a, b] = c

adj_mat = csr_matrix(adj_mat, dtype='int')
D = dijkstra(adj_mat, directed=False)
# print(D)

# 最短経路じゃない辺全探索
edge_not_use = set()
# for s, t in combinations(range(N), 2):
#     d = D[s, t]
#     for (i, j), e in edges.items():
#         if d != D[s, i] + e + D[j, t] and d != D[s, j] + e + D[i, t]:  # 最短経路辺になれない
#             print(s, i, j, t)
#             print(d, '|', D[s, i], e, D[j, t], '|',
#                   D[s, j], e, D[i, t])  # 最短経路辺になれない
#             edge_not_use.add((i, j) if i < j else (j, i))
#     print()

# 辺を固定してs,tを全探索したほうが良さそう
for (i, j), e in edges.items():
    for s, t in combinations(range(N), 2):
        d = D[s, t]
        if d == D[s, i] + e + D[j, t] or d == D[t, i] + e + D[j, s]:
            break
    else:
        edge_not_use.add((i, j))


print(len(edge_not_use))
# print(edge_not_use)