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)

#!/usr/bin/env python
# -*- coding: utf-8 -*-
#
# FileName: 	D_fix
# CreatedDate:  2020-02-21 01:20:19 +0900
# LastModified: 2020-02-21 02:04:42 +0900
#


import os
import sys
import numpy as np
import heapq
#import pandas as pd
def make_matrix(matrix,a,b,c):
    matrix[a-1][b-1] = c
    matrix[b-1][a-1] = c

def initial_signal(i,matrix):
    d = [100]*(matrix.shape[0])
    d[i] = 0
    Q = []
    heapq.heappush(Q,(0,i))
    return d,Q

def relax(matrix,u,v,d,Q):
    if d[v] > d[u]+matrix[u][v]:
        d[v] = d[u]+matrix[u][v]
        heapq.heappush(Q, (d[v],v))


def dijkstra(i,matrix):
    d,Q = initial_signal(i,matrix)
    while Q:
        cost,u = heapq.heappop(Q)
        if d[u]<cost:
            continue
        for v in range(matrix.shape[0]):
            if matrix[u][v]!=0:
                relax(matrix,u,v,d,Q)
    return d
    

def main():
    N,M = map(int, input().split())
    matrix = np.zeros((N,N),dtype=np.int32)
    for _ in range(M):
        a,b,c = map(int,input().split())
        make_matrix(matrix,a,b,c)
    shortest = []
    for i in range(matrix.shape[0]):
        short = dijkstra(i,matrix)
        shortest.append(short)
    ans=0
    for i in range(N):
        for j in range(N):
            if matrix[i][j]!=0 and matrix[i][j]>shortest[i][j]:
                ans+=1

    print(ans//2)


if __name__ == "__main__":
    main()