/*
    2874    1731MS    29448K    2286 B    G++     2014300227
*/
#include <bits/stdc++.h>
using namespace std;
const int N=1e4+4;
typedef long long ll;
const double PI=acos(-1.0);
int n,m,c,cnt,head[N],a,b,num,pre[N],p[N],l,r,v,dis[N],ans[1000005],vis[N],fa[N];
struct Edge
{
    int to,va,next;
};
Edge edge[2*N];
struct ques
{
    int to,next,id;
};
ques que[2000000+20];
void add(int s,int e,int v)
{
    //edge[cnt].fr = s;
    edge[cnt].to = e;
    edge[cnt].va=v;
    edge[cnt].next = head[s];
    head[s] = cnt++;//学会了这种存边的方法;
}
void q_add(int s,int e,int order)
{
    //que[num].fr = s;
    que[num].to = e;
    que[num].next = pre[s];
    que[num].id=order;
    pre[s] = num++;
}
int findset(int x)
{
    if(x == p[x])return x;
    return p[x] = findset(p[x]);
}
int fun(int x)
{
    if(x==fa[x])return x;
    return fa[x]=fun(fa[x]);
}
void Tarjan(int x,int dist)
{
    vis[x] = 1;
    dis[x]=dist;
    for(int i = head[x];i!=-1;i = edge[i].next)//head[x]指向以x为端点的一条边;下面的pre[x]也是相同的道理;
    {
        int y = edge[i].to;
        if(!vis[y])
        {
            Tarjan(y,dist+edge[i].va);
            fa[y] = x;
        }
    }//前边表示这个节点的所有子树已经处理完毕,下面可以回答相关的询问了;
    for(int i = pre[x];i!=-1;i = que[i].next)
    {
        int y = que[i].to;
        if(findset(x) == findset(y))
        {
            if(vis[y])
            ans[que[i].id] = dis[y]+dis[x]-2*dis[fun(y)];
        }
        else ans[que[i].id] = -1;
    }
}
int main()
{
    while(scanf("%d%d%d",&n,&m,&c)!=EOF)
    {
        for(int i = 0;i <= n;i++)
        {
            p[i] = i;
            fa[i] = i;
            head[i]=pre[i]=-1;
            vis[i]=0;
        }
        cnt = 0;
        num = 0;
        for(int i = 0;i < m;i++)
        {
            scanf("%d%d%d",&l,&r,&v);
            int fx=findset(l),fy=findset(r);
            if(fx!=fy)p[fx]=fy;//看是否在一个树上的并查集
            add(l,r,v);
            add(r,l,v);
        }
        for(int i = 0;i < c;i++)
        {
            scanf("%d%d",&a,&b);
            q_add(a,b,i);
            q_add(b,a,i);
        }
        for(int i=1;i<=n;i++)
        {
            if(!vis[i])
            {
                Tarjan(i,0);
            }
        }
        for(int i = 0;i < c;i++)
        {
            if(ans[i]>-1)printf("%d\n",ans[i]);
            else printf("Not connected\n");
        }
    }
    return 0;
}