/*
Solution for APIO 2016 - Merchant
Tags : binary search, shortest path (Floyd-Warshall)
Complexity O(n^3 * log (MOD))
*/
#include <bits/stdc++.h>
using namespace std;
#pragma GCC target ("avx2")
#pragma GCC optimization ("Ofast")
#pragma GCC optimization ("unroll-loops")
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef long double ld;
typedef pair<ll, ll> pll;
#define FOR(i, a, b) for(int i = a; i < b; i++)
#define ROF(i, a, b) for(int i = a; i >= b; i--)
#define ms memset
#define pb push_back
#define fi first
#define se second
ll MOD = 1000000007;
ll MOD1 = 998244353;
ll power(ll base, ll n){
if (n == 0) return 1;
if (n == 1) return base;
ll halfn = power(base, n/2);
if (n % 2 == 0) return (halfn * halfn) % MOD;
return (((halfn * halfn) % MOD) * base) % MOD;
}
ll inverse(ll n){
return power(n, MOD-2);
}
ll add(ll a, ll b){
return (a+b) % MOD;
}
ll mul(ll a, ll b){
a %= MOD;
return (a*b) % MOD;
}
ll gcd(ll a, ll b){
if (a == 1) return 1;
if (a == 0) return b;
return gcd(b%a, a);
}
ld pi = 3.141592653589793238;
void solve(){
int n, m, k; cin >> n >> m >> k;
ll b[n+1][k], s[n+1][k];
FOR(i,1,n+1){
FOR(j,0,k){
cin >> b[i][j] >> s[i][j];
//b[i][j] = buying price for jth item in ith market
//s[i][j] = selling price
}
}
ll mat[n+1][n+1];
ms(mat,-1,sizeof(mat));
FOR(i,0,m){
int u, v, w; cin >> u >> v >> w;
mat[u][v] = w;
}
FOR(i,1,n+1){
FOR(j,1,n+1){
if (mat[i][j]==-1) mat[i][j] = MOD;
}
}
// O(n^3) Floyd-Warshall to find shortest distance between any 2 markets
FOR(i,1,n+1){
FOR(j,1,n+1){
FOR(l,1,n+1){
mat[j][l] = min(mat[j][l], mat[j][i]+mat[i][l]);
}
}
}
ll profit[n+1][n+1];
FOR(i,1,n+1){
FOR(j,1,n+1){
ll mx = 0; // not buying/selling item
FOR(l,0,k){
if (b[i][l] != -1 && s[j][l] != -1) mx = max(mx, s[j][l] - b[i][l]);
}
profit[i][j] = mx;
// max profit from buying an item in market i and selling it in market j
}
}
ll L = 0, R = MOD;
while (R-L>1){
ll M = (L+R)/2;
bool sol = false;
// does there exist a nonnegative positive cycle?
ll dist[n+1][n+1];
FOR(i,1,n+1){
FOR(j,1,n+1){
if (mat[i][j] == MOD) dist[i][j] = -MOD;
else dist[i][j] = profit[i][j] - M*mat[i][j];
}
}
FOR(i,1,n+1){
FOR(j,1,n+1){
FOR(l,1,n+1){
dist[j][l] = max(dist[j][l], dist[j][i] + dist[i][l]);
}
}
}
FOR(i,1,n+1){
if (dist[i][i] >= 0){
sol = true; break;
}
}
if (sol) L=M;
else R=M;
}
cout << L << '\n';
return;
}
int main() {
cout << fixed << setprecision(8);
ios::sync_with_stdio(false);
if (fopen("input.txt", "r")) {
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
}
int TC = 1;
//cin >> TC;
FOR(i, 1, TC+1){
//cout << "Case #" << i << ": ";
solve();
}
}
