//Floyd不可以处理负权回路, 用spfa判断是否有负权回路
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 210, INF = 1e9;

int n, m, q;
int d[N][N];//d[i][j] 结点i到结点j的最短距离

void Floyd()
{
    for (int k = 1; k <= n; k ++ )
        for (int i = 1; i <= n; i ++ )
            for (int j = 1; j <= n; j ++ )
                d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
}

int main()
{
    scanf("%d%d%d", &n, &m, &q);

    //初始化距离，为下面的防止自环做铺垫！
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= n; j ++ )
            if (i == j) d[i][j] = 0;//自己到自己的距离为0,防止了自环
            else d[i][j] = INF;//否则无穷大

    for (int i = 0; i < m; i ++ )
    {
        int a, b, w;
        scanf("%d%d%d", &a, &b, &w);//读入结点a到结点b的距离w

        d[a][b] = min(d[a][b], w);//有两个作用，实在是妙
        //若存在自环，必然是正数的，大于0，就更新不了
        //若存在重边，则选择距离最短的一条
    }

    Floyd();

    for (int i = 0; i < q; i ++ )
    {
        int a, b;
        scanf("%d%d", &a, &b);

        if (d[a][b] > INF / 2) puts("impossible");
        else printf("%d\n", d[a][b]);
    }

    return 0;
}