#include<cstdio>

 #include<iostream>

 #include<algorithm>

 #include<cstring>

 #include<cmath>

 #include<queue>

 #include<map>

 using namespace std;

 #define MOD 1000000007

 const int INF=0x3f3f3f3f;

 const double eps=1e-;

 typedef long long ll;

 #define cl(a) memset(a,0,sizeof(a))

 #define ts printf("*****\n");

 int n,m,tt;

 /*

 * 求 无向图的割点和桥

 * 可以找出割点和桥，求删掉每个点后增加的连通块。

 * 需要注意重边的处理，可以先用矩阵存，再转邻接表，或者进行判重

 */

 const int MAXN = ;

 const int MAXM = ;

 struct Edge

 {

     int to,next;

     int w;

     bool cut;//是否为桥的标记

 }edge[MAXM];

 int head[MAXN],tot;

 int Low[MAXN],DFN[MAXN],Stack[MAXN];

 int Index,top;

 bool Instack[MAXN];

 bool cut[MAXN];

 int add_block[MAXN];//删除一个点后增加的连通块

 int bridge;

 void addedge(int u,int v,int w)

 {

     edge[tot].to = v;edge[tot].next = head[u];edge[tot].cut = false;

     edge[tot].w = w;

     head[u] = tot++;

 }

 void Tarjan(int u,int pre)

 {

     int v;

     Low[u] = DFN[u] = ++Index;

     Stack[top++] = u;

     Instack[u] = true;

     int son = ;

     int pre_num=;

     for(int i = head[u];i != -;i = edge[i].next)

     {

         v = edge[i].to;

         if(v == pre&&pre_num==)

         {

             pre_num++;

             continue;

         }

         if( !DFN[v] )

         {

             son++;

             Tarjan(v,u);

             if(Low[u] > Low[v])Low[u] = Low[v];

             //桥

             //一条无向边(u,v)是桥，当且仅当(u,v)为树枝边，且满足DFS(u)<Low(v)。

             if(Low[v] > DFN[u])

             {

                 bridge++;

                 edge[i].cut = true;

                 edge[i^].cut = true;

             }

             //割点

             //一个顶点u是割点，当且仅当满足(1)或(2) (1) u为树根，且u有多于一个子树。

             //(2) u不为树根，且满足存在(u,v)为树枝边(或称父子边，

             //即u为v在搜索树中的父亲)，使得DFS(u)<=Low(v)

             if(u != pre && Low[v] >= DFN[u])//不是树根

             {

                 cut[u] = true;

                 add_block[u]++;

             }

         }

         else if( Low[u] > DFN[v])

         Low[u] = DFN[v];

     }

     //树根，分支数大于1

     if(u == pre && son > )cut[u] = true;

     if(u == pre)add_block[u] = son - ;

     Instack[u] = false;

     top--;

 }

 int solve(int N)

 {

     memset(DFN,,sizeof(DFN));

     memset(Instack,false,sizeof(Instack));

     memset(add_block,,sizeof(add_block));

     memset(cut,false,sizeof(cut));

     Index = top = ;

     bridge = ;

     for(int i = ;i <= N;i++)

         if(!DFN[i])

             Tarjan(i,i);

     int ret=INF;

     for(int u = ;u <= N;u++)

     for(int i = head[u];i != -;i = edge[i].next)

     if(edge[i].cut)

     {

         ret=min(ret,edge[i].w);

     }

     if(ret==INF)    return -;

     if(ret==)  ret++;

     return ret;

 }

 int F[MAXN];

 int find(int x)

 {

     if(F[x] == -)return x;

     else return F[x] = find(F[x]);

 }

 void init()

 {

     memset(F,-,sizeof(F));

     tot = ;

     memset(head,-,sizeof(head));

 }

 void bing(int u,int v)

 {

     int t1 = find(u);

     int t2 = find(v);

     if(t1 != t2)F[t1] = t2;

 }

 int main()

 {

     int n,m;

     #ifndef ONLINE_JUDGE

     freopen("1.in","r",stdin);

     #endif

     while(scanf("%d%d",&n,&m) == )

     {

         if(n ==  && m == )break;

         int u,v,w;

         init();

         while(m--)

         {

             scanf("%d%d%d",&u,&v,&w);

             if(u == v)continue;

             addedge(u,v,w);

             addedge(v,u,w);

             bing(u,v);

         }

         bool flag = true;

         for(int i = ; i <= n;i++)

             if(find(i) != find())

                 flag = false;

         if(!flag)

         {

             printf("0\n");

             continue;

         }

         printf("%d\n",solve(n));

     }

     return ;

 }