uva1395

/*
考察点：kruskal算法生成最小生成树
题目实现步骤：
1：把边权值从大到小排列，使用kruskal算法生成最小树
2：假定最小树中的最小边为min,最大边为max,则这棵生成树包含的边集合为(min, max)
3：所以苗条度为max - min
4：但这样的苗条度不一定最小，所以对整张图的边的集合，枚举min,然后从min开始kruskal算法
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>

using namespace std;
const int maxn = 5000 +5;
const int INF = 999999;
int n, m, p[maxn], ch[maxn];

struct Side {
int u, v, val;
} sides[maxn];

bool cmp(const Side& Side1, const Side& Side2) {
return Side1.val < Side2.val;
}

int UFind(int u) {
return p[u] == u ? u : UFind(p[u]);
}

int solve() {   //(L,R)
int slimness = INF;
for(int dex = 0; dex < m; dex++) {  //外围枚举L
int sideNum = 0;

for(int i = 1; i <= n; i++) {    //并查集初始化,注意节点编号从1开始
p[i] = i;
}

for(int i = dex; i < m; i++) {
int u = UFind(sides[i].u), v = UFind(sides[i].v);
if(u != v) {
p[u] = v;
if(++sideNum == n - 1)  {slimness = min(sides[i].val - sides[dex].val, slimness);   break;}
}
}
}
if(slimness == INF)     slimness = -1;
return slimness;
}

int main()
{
//freopen("input.txt", "r", stdin);
while(scanf("%d%d", &n, &m) == 2 && n) {
for(int i = 0; i < m; i++)
scanf("%d%d%d", &sides[i].u, &sides[i].v, &sides[i].val);

sort(sides, sides + m, cmp);

if (m < n - 1)  printf("-1\n");
else                printf("%d\n", solve());
}
return 0;
}