// Om Namah Shivaya
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x, y) ((x + y - 1) / (y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
#define rep(i, n) for(int i = 0; i < n; ++i)
#define rep1(i, n) for(int i = 1; i <= n; ++i)
#define rev(i, s, e) for(int i = s; i >= e; --i)
#define trav(i, a) for(auto &i : a)
template<typename T>
void amin(T &a, T b) {
a = min(a, b);
}
template<typename T>
void amax(T &a, T b) {
a = max(a, b);
}
#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif
/*
refs:
https://usaco.guide/problems/apio-2017traveling-merchant/solution
profit = cost/length
lets say profit = cost (we ignore length for now)
how to solve for this case?
for every pair (u,v), create an edge if v is reachable from u
this graph says: if we are @u with an empty bag, we buy something @u, then sell it @v, what's the max profit?
the weights of the edges denote this profit
(we can also choose to buy nothing, weight = 0)
find the max cost cycle (inf possible)
come back to original problem
profit = cost/length
we also consider length
so when buying something from u and selling it @v, it's optimal to take the shortest path from u to v
so for every pair (u,v), we compute:
(max_add, shortest_path)
profit function is a ratio, annoying to deal with
think b.s in such cases
how to check if profit >= mid?
cost/length >= mid
cost >= mid*length
cost - mid*length >= 0
edge (max_add, shortest_path) becomes:
edge with weight (max_add - mid*shortest_path)
find if there is a cycle with sum of weights >= 0
=> floyd-warshalls
*/
const int MOD = 1e9 + 7;
const int N = 100 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;
const int K = 1e3 + 5;
ll buy[N][K], sell[N][K];
ll sp[N][N], best_buy_sell[N][N];
ll weight[N][N], lp[N][N]; // lp = longest path
void solve(int test_case)
{
ll n,m,k; cin >> n >> m >> k;
rep1(i,n){
rep1(j,k){
cin >> buy[i][j];
}
rep1(j,k){
cin >> sell[i][j];
}
}
memset(sp,0x10,sizeof sp);
rep1(i,m){
ll u,v,w; cin >> u >> v >> w;
sp[u][v] = w;
}
rep1(p,n){
rep1(i,n){
rep1(j,n){
amin(sp[i][j], sp[i][p] + sp[p][j]);
}
}
}
rep1(i,n){
rep1(j,n){
if(i == j) conts;
if(sp[i][j] >= inf2) conts;
rep1(p,k){
if(buy[i][p] != -1 and sell[j][p] != -1){
ll val = sell[j][p] - buy[i][p];
amax(best_buy_sell[i][j], val);
}
}
}
}
auto ok = [&](ll mid){
memset(weight,-0x10,sizeof weight);
memset(lp,-0x10,sizeof lp);
rep1(i,n){
rep1(j,n){
if(i == j) conts;
if(sp[i][j] >= inf2) conts;
ll w = best_buy_sell[i][j] - mid*sp[i][j];
weight[i][j] = lp[i][j] = w;
}
}
rep1(p,n){
rep1(i,n){
rep1(j,n){
amax(lp[i][j], lp[i][p] + lp[p][j]);
}
}
}
rep1(i,n){
rep1(j,n){
if(i == j) conts;
ll w = lp[i][j] + lp[j][i];
if(w >= 0){
return true;
}
}
}
return false;
};
ll l = 0, r = inf1;
ll ans = -inf2;
while(l <= r){
ll mid = (l+r) >> 1;
if(ok(mid)){
ans = mid;
l = mid+1;
}
else{
r = mid-1;
}
}
assert(ans != -inf2);
cout << ans << endl;
}
int main()
{
fastio;
int t = 1;
// cin >> t;
rep1(i, t) {
solve(i);
}
return 0;
}
