Problem:
Problem F: Slim Span

Given an undirected weighted graph G, you should find one of spanning trees specified as follows.

The graph G is an ordered pair (V, E), where V is a set of vertices {v1, v2, ... , vn} and E is a set of undirected edges {e1, e2, ... , em}. Each edge e ∈ E has its weight w(e).

A spanning tree T is a tree (a connected subgraph without cycles) which connects all the n
vertices with n - 1 edges. The slimness of a spanning tree T is defined as the difference between
the largest weight and the smallest weight among the n - 1 edges of T.

Figure 5: A graph G and the weights of the edges

For example, a graph G in Figure 5(a) has four vertices {v1, v2, v3, v4} and five undirected edges {e1, e2, e3, e4, e5}. The weights of the edges are w(e1) = 3, w(e2) = 5, w(e3) = 6, w(e4) = 6, w(e5) = 7 as shown in Figure 5(b).

Figure 6: Examples of the spanning trees of G

There are several spanning trees for G. Four of them are depicted in Figure 6(a)-(d). The spanning tree Ta in Figure 6(a) has three edges whose weights are 3, 6 and 7. The largest weight
is 7 and the smallest weight is 3 so that the slimness of the tree Ta is 4. The slimnesses of
spanning trees Tb , Tc and Td shown in Figure 6(b), (c) and (d) are 3, 2 and 1, respectively. You
can easily see the slimness of any other spanning tree is greater than or equal to 1, thus the
spanning tree Td in Figure 6(d) is one of the slimmest spanning trees whose slimness is 1.

Your job is to write a program that computes the smallest slimness.

Input

The input consists of multiple datasets, followed by a line containing two zeros separated by a space. Each dataset has the following format.

n m
a1 b1 w1
.
.
.
am bm wm

Every input item in a dataset is a non-negative integer. Items in a line are separated by a space.

n is the number of the vertices and m the number of the edges. You can assume 2 ≤ n ≤ 100 and 0 ≤ m ≤ n(n - 1)/2. ak and bk (k = 1, ... , m) are positive integers less than or equal to n, which represent the two vertices vak and vbk connected by the kth edge ek. wk is a positive integer less than or equal to 10000, which indicates the weight of ek . You can assume that the graph G = (V, E) is simple, that is, there are no self-loops (that connect the same vertex) nor parallel edges (that are two or more edges whose both ends are the same two vertices).

Output

For each dataset, if the graph has spanning trees, the smallest slimness among them should be
printed. Otherwise, -1 should be printed. An output should not contain extra characters.

Sample Input

4 5
1 2 3
1 3 5
1 4 6
2 4 6
3 4 7
4 6
1 2 10
1 3 100
1 4 90
2 3 20
2 4 80
3 4 40
2 1
1 2 1
3 0
3 1
1 2 1
3 3
1 2 2
2 3 5
1 3 6
5 10
1 2 110
1 3 120
1 4 130
1 5 120
2 3 110
2 4 120
2 5 130
3 4 120
3 5 110
4 5 120
5 10
1 2 9384
1 3 887
1 4 2778
1 5 6916
2 3 7794
2 4 8336
2 5 5387
3 4 493
3 5 6650
4 5 1422
5 8
1 2 1
2 3 100
3 4 100
4 5 100
1 5 50
2 5 50
3 5 50
4 1 150
0 0

Output for the Sample Input

1
20
0
-1
-1
1
0
1686
50

Solution in C (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#define inf (int)(1e9)

typedef struct {
	int N;
	int *u;
	int *u_rank;
}union_find;

union_find *make_union_find(int N){
	int i;
	union_find *uf = (union_find *)malloc(sizeof(union_find));
	uf->N = N;
	uf->u = (int *)malloc(sizeof(int) * N);
	uf->u_rank = (int *)malloc(sizeof(int) * N);
	for(i = 0; i < N; i++){
		(uf->u)[i] = i;
		(uf->u_rank)[i] = 1;
	}
	return uf;
}

int root_uf(int x, union_find *uf){
	int *u = uf->u;
	if(u[x] == x){
		return x;
	}
	else{
		u[x] = root_uf(u[x], uf);
		return u[x];
	}
}

void combine_uf(int x, int y, union_find *uf){
	int x_root = root_uf(x, uf);
	int y_root = root_uf(y, uf);
	int *u = uf->u;
	int *u_rank = uf->u_rank;
	if(x_root == y_root){
		return;
	}
	else if(u_rank[x_root] < u_rank[y_root]){
		u[x_root] = y_root;
		u_rank[y_root] += u_rank[x_root];
		u_rank[x_root] = 0;
	}
	else{
		u[y_root] = x_root;
		u_rank[x_root] += u_rank[y_root];
		u_rank[y_root] = 0;
	}
}

//xとyが同じ集合に属していれば1を,そうでなければ0を返す
int is_same_union_uf(int x, int y, union_find *uf){
	if(root_uf(x, uf) == root_uf(y, uf)){
		return 1;
	}
	else{
		return 0;
	}
}

//xが属する集合の要素数を返す
int rank_uf(int x, union_find *uf){
	return (uf->u_rank)[root_uf(x, uf)];
}

typedef struct {
	int a;
	int b;
	int w;
}edge;

int min(int a, int b){
	return a <= b ? a : b;
}

edge es[5000];

int compare(const void *a, const void *b){
	edge A = *(edge *)a, B = *(edge *)b;
	return A.w - B.w;
}

int main(){
	int N, M, i, j;
	while(1){
		scanf("%d%d", &N, &M);
		if(N == 0 && M == 0){
			return 0;
		}
		for(i = 0; i < M; i++){
			scanf("%d%d%d", &es[i].a, &es[i].b, &es[i].w);
			es[i].a--;
			es[i].b--;
		}
		qsort(es, M, sizeof(edge), compare);
		int ans = inf;
		for(i = 0; i < M; i++){
			union_find *uf = make_union_find(N);
			for(j = i; j < M; j++){
				combine_uf(es[j].a, es[j].b, uf);
				if(rank_uf(0, uf) == N){
					ans = min(ans, es[j].w - es[i].w);
					break;
				}
			}
		}
		if(ans == inf){
			printf("-1\n");
		}
		else{
			printf("%d\n", ans);
		}
	}
}