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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<string.h>
#define FOR(i,a,b) for(int i=a;i<b;i++)
#define INF (int)1e9
#define MIN(a,b) ((a)<(b) ? (a):(b))
int main(){
	int n,m,a[1002],b[1002],c[1002],root[102][102];

	memset(root,INF,sizeof(root));
	FOR(i,0,102) root[i][i] = 0;

	scanf("%d %d",&n,&m);
	FOR(i,0,m){
		scanf("%d %d %d",&a[i],&b[i],&c[i]);
		root[a[i]][b[i]] = c[i];
		root[b[i]][a[i]] = c[i];
	}

	FOR(k,1,n+1)	//中継点
		FOR(i,1,n+1)	//始点
			FOR(j,1,n+1)	//終点
				root[i][j] = MIN(root[i][j],root[i][k]+root[k][j]);

	int ans = m;

	FOR(k,0,m){	//判定する辺
		FOR(i,1,n+1){	//始点
			FOR(j,1,n+1){	//終点
				if(root[i][j] == root[i][a[k]] + c[k] + root[b[k]][j] ||
						root[i][j] == root[j][a[k]] + c[k] + root[b[k]][i]){
					ans--;
					goto NEXT;
				}
			}
		}
		NEXT:;
	}



//			FOR(i,1,n+1)	//始点
//				if(root[i][a[k]] + c[k] == root[i][b[k]]){	//最短経路に使用済み
//					ans--;
//					break;
//				}



	printf("%d\n",ans);
}