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 Lua (5.3.2) (judge verdict: Accepted)

local n, m = io.read("*n", "*n")
local edge = {}
local edgeidx = {}
for i = 1, n do
  edge[i] = {}
  edgeidx[i] = {}
end
local edgeuse = {}
for i = 1, m do
  edgeuse[i] = false
  local a, b, c = io.read("*n", "*n", "*n")
  edge[a][b] = c
  edgeidx[a][b] = i
  edge[b][a] = c
  edgeidx[b][a] = i
end
local inf = 1000000007
local len = {}
local asked = {}
local function getpos()
  local cand = 0
  for i = 1, n do
    if not asked[i] then
      if cand == 0 or len[i] < len[cand] then
        cand = i
      end
    end
  end
  return cand
end
local function solve(spos)
  for i = 1, n do
    asked[i] = false
    len[i] = i == spos and 0 or inf
  end
  for i = 1, n do
    local src = getpos()
    asked[src] = true
    for dst, cost in pairs(edge[src]) do
      if len[dst] + cost == len[src] then
        edgeuse[edgeidx[dst][src]] = true
      end
      if len[src] + cost < len[dst] then
        len[dst] = len[src] + cost
      end
    end
  end
end

for i = 1, n do
  solve(i)
end

local ret = 0
for i = 1, m do
  if not edgeuse[i] then ret = ret + 1 end
end
print(ret)