/*
题目大意：
从a点到b点， 找到一条路径，使得这条路径上的所有噪音中最大的值是所有路径中最小的，
这个噪音值便是要求的。若不连通，输出no path
方法1：floyd算法，改变map[i][j]的含义，改变松弛条件：
map[i][j] = min(map[i][j], max(map[i][k], map[k][j]));
方法2：kruscal算法，先对边排序，然后从小到大加边。如果start与endd相连，更新ans。
*/
/*
#include <cstdio>
#include <algorithm>
#define N 105
using namespace std;
const int inf = 0x3f3f3f3f;
int map[N][N];
int n, m, q;
void floyd()
{
for (int k = 1; k <= n; k++)
{
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
{
if (map[i][k] < inf && map[k][j] < inf)
map[i][j] = min(map[i][j], max(map[i][k], map[k][j]));
}
}
}
}
int main()
{
int cas = 0;
while (scanf("%d%d%d", &n, &m, &q) != EOF && (n || m || q))
{
if (cas) printf("\n");
printf("Case #%d\n", ++cas);
int t1, t2, t3;
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= n; j++)
{
if (i == j) map[i][j] = 0;
else map[i][j] = inf;
}
}
for (int i = 1; i <= m; i++)
{
scanf("%d%d%d", &t1, &t2, &t3);
if(map[t1][t2] > t3)
map[t1][t2] = map[t2][t1] = t3;
}
floyd();
int start, endd;
while (q--)
{
scanf("%d%d", &start, &endd);
if (map[start][endd] >= inf) printf("no path\n");
else printf("%d\n", map[start][endd]);
}
}
return 0;
}
*/
#include <cstdio>
#include <algorithm>
#define N 105
using namespace std;
const int inf = 0x3f3f3f3f;
typedef struct note {
int u; int v; int w;
}edge;
edge e[N * 10];
int n, m, q;
int f[N];
void init()
{
for (int i = 1; i <= n; i++)
f[i] = i;
}
int find(int x)
{
return x == f[x] ? x : f[x] = find(f[x]);
}
int merge(int x, int y)
{
int t1, t2;
t1 = find(x); t2 = find(y);
if (t1 != t2) { f[t2] = t1; return 1; }
return 0;
}
int cmp(edge a, edge b)
{
return a.w < b.w;
}
int main()
{
int cas = 0;
while (scanf("%d%d%d", &n, &m, &q) != EOF && (n || m || q))
{
for (int i = 0; i < m; i++)
scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w);
sort(e, e + m, cmp);
int start, endd, ans;
if (cas) printf("\n");
printf("Case #%d\n", ++cas);
while (q--)
{
ans = inf;
scanf("%d%d", &start, &endd);
init();
for (int i = 0; i < m; i++)
{
merge(e[i].u, e[i].v);
if (find(start) == find(endd)) { ans = e[i].w; break; }
}
if (ans == inf) printf("no path\n");
else printf("%d\n", ans);
}
}
}
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