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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define  rep(i, n) for(int i = 0; i < n; i++) 
#define  all(x) (x).begin(),(x).end()     // 昇順ソート
#define  rall(v) (v).rbegin(), (v).rend() // 降順ソート
#define  INF 1LL << 60
typedef long long int LL;
typedef long long int ll;
#define pll pair<ll, ll>
#define F first
#define S second
const int MOD = 1000000007;
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return true; } return false; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return true; } return false; }
//sort(all(x))とするとソートできるよ
// 10^x は pow(10,(x)) 
// 任意のlogは　対数の底の変換を使う    log(N) / log(任意の底)

void WF(int n,vector<vector<LL>> &d) {
  for (int k = 0; k < n; k++){       // 経由する頂点
    for (int i = 0; i < n; i++) {    // 始点
      for (int j = 0; j < n; j++) {  // 終点
        d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
      }
    }
  }
}

int main(){
    int N,M;cin >> N >> M;
    vector<int> a(M);
    vector<int> b(M);
    vector<int> c(M);

    vector<vector<LL>> G(N,vector<LL>(N,1 << 30));

    rep(i,M){
        cin >> a[i] >> b[i] >> c[i];
        a[i]--;b[i]--;
        G[a[i]][b[i]] = c[i];
        G[b[i]][a[i]] = c[i];
    }

    WF(N,G);
    int ans = 0;
    rep(i,M){
        if(G[a[i]][b[i]] < c[i])ans++;
    }
    cout << ans << endl;
}