#include<iostream>#include<cstring>#include<set>#include<map>#include<cmath>#include<stack>#include<queue>#include<deque>#include<list>#include<algorithm>#include<stdio.h>#include<iomanip>#define rep(i,n) for(int i=0;i<n;++i)#define fab(i,a,b) for(int i=a;i<=b;++i)#define fba(i,b,a) for(int i=b;i>=a;--i)#define PB push_back#define INF 0x3f3f3f3f#define MP make_pair#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define sf scanf#define pf printf#define LL long longconst int N=100010;const int maxn=N;const int logmaxn=20;using namespace std;typedef pair<int,int>PII;/*struct LCA {  int n;  int fa[maxn];   // 父亲数组  int cost[maxn]; // 和父亲的费用  int L[maxn];    // 层次（根节点层次为0）  int anc[maxn][logmaxn];     // anc[p][i]是结点p的第2^i级父亲。anc[i][0] = fa[i]  int maxcost[maxn][logmaxn]; // maxcost[p][i]是i和anc[p][i]的路径上的最大费用  // 预处理，根据fa和cost数组求出anc和maxcost数组  void preprocess() {    for(int i = 0; i < n; i++) {      anc[i][0] = fa[i]; maxcost[i][0] = cost[i];      for(int j = 1; (1 << j) < n; j++) anc[i][j] = -1;    }       for(int j = 1; (1 << j) < n; j++)      for(int i = 0; i < n; i++)        if(anc[i][j-1] != -1) {          int a = anc[i][j-1];          anc[i][j] = anc[a][j-1];          maxcost[i][j] = max(maxcost[i][j-1], maxcost[a][j-1]);        }  }  // 求p到q的路径上的最大权  int query(int p, int q) {    int tmp, log, i;    if(L[p] < L[q]) swap(p, q);    for(log = 1; (1 << log) <= L[p]; log++); log--;    int ans = -INF;    for(int i = log; i >= 0; i--)      if (L[p] - (1 << i) >= L[q]) { ans = max(ans, maxcost[p][i]); p = anc[p][i];}       if (p == q) return ans; // LCA为p       for(int i = log; i >= 0; i--)      if(anc[p][i] != -1 && anc[p][i] != anc[q][i]) {        ans = max(ans, maxcost[p][i]); p = anc[p][i];        ans = max(ans, maxcost[q][i]); q = anc[q][i];      }    ans = max(ans, cost[p]);    ans = max(ans, cost[q]);    return ans; // LCA为fa[p]（它也等于fa[q]）  }  };*/struct LCA{   int n,fa[N],cost[N],L[N],anc[N][20],maxcost[N][20];   void preprocess(){       rep(i,n){           anc[i][0]=fa[i];maxcost[i][0]=cost[i];           for(int j=1;(1<<j)<n;j++){               anc[i][j]=-1;           }       }       for(int j=1;(1<<j)<n;j++){           rep(i,n){               if(anc[i][j-1]!=-1){                   int p=anc[i][j-1];                   anc[i][j]=anc[p][j-1];                   maxcost[i][j]=max(maxcost[i][j-1],maxcost[p][j-1]);               }           }       }   }   int query(int p,int q){        int k=0;        if(L[p]<L[q])swap(p,q);        while((1<<(k+1))<=L[p])k++;   //     for(k=1;(1<<k)<=L[p];k++);k--;        int ans=-INF;        fba(i,k,0)if(L[p]-(1<<i)>=L[q]){            ans=max(ans,maxcost[p][i]);            p=anc[p][i];        }        if(p==q)return ans;/// lca = p         fba(i,k,0)if(anc[p][i]!=-1&&anc[p][i]!=anc[q][i]){            ans=max(ans,maxcost[p][i]);p=anc[p][i];            ans=max(ans,maxcost[q][i]);q=anc[q][i];        }        ans=max(ans,cost[p]);        ans=max(ans,cost[q]);        return ans;// lca = fa[p] = fa[q];   }};LCA solver;int pa[N];int find(int x){    return x==pa[x]?x:pa[x]=find(pa[x]);}vector<PII>G[N];void dfs(int u,int fa,int lev){    solver.L[u]=lev;    rep(i,G[u].size()){        int v=G[u][i].first;        if(v!=fa){            solver.fa[v]=u;            solver.cost[v]=G[u][i].second;            dfs(v,u,lev+1);        }    }}struct Edge{    int u,v,d;    bool operator< (const Edge& rhs)const{        return d < rhs.d;    }}E[N];int main(){    int cas=0,n,m,Q;    while(~sf("%d%d",&n,&m)&&n){        rep(i,m){            sf("%d%d%d",&E[i].u,&E[i].v,&E[i].d);            E[i].u--;E[i].v--;        }        sort(E,E+m);        rep(i,n){pa[i]=i;G[i].clear();}        rep(i,m){            int x=E[i].u,y=E[i].v,u=find(x),v=find(y);            if(u!=v){                pa[u]=v;                G[x].PB(MP(y,E[i].d));                G[y].PB(MP(x,E[i].d));            }        }        solver.n=n;        dfs(0,-1,0);        solver.preprocess();        if(++cas!=1)pf("\n");        sf("%d",&Q);        while(Q--){            int a,b;sf("%d%d",&a,&b);            pf("%d\n",solver.query(a-1,b-1));        }    }    return 0;}