// 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:
edi
let's look at the initial array
when will some guy definitely not be in the best triplet?
=> if he dominates in >= 2 areas
so we can remove all guys who cant be there
repeat the process on the resulting array
repeat until we can no longer delete anyone
what does it mean if we can make no more deletions?
there is exactly 1 guy dominating each area
so we can pick the guy dominating each of the 3 areas and that would be our triplet
if < 3 guys are remaining, then not possible
naive simulation of this process = O(n^2)
*/
const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;
void solve(int test_case)
{
ll n; cin >> n;
vector<array<ll,3>> a(n);
rep(i,n) rep(j,3) cin >> a[i][j];
while(!a.empty()){
array<ll,3> best;
best.fill(0);
trav(ar,a){
rep(j,3){
amax(best[j], ar[j]);
}
}
vector<array<ll,3>> b;
trav(ar,a){
ll cnt = 0;
rep(j,3){
if(ar[j] == best[j]){
cnt++;
}
}
if(cnt <= 1){
b.pb(ar);
}
}
if(sz(a) == sz(b)){
ll ans = 0;
rep(j,3) ans += best[j];
cout << ans << endl;
return;
}
a = b;
}
cout << -1 << endl;
}
int main()
{
fastio;
int t = 1;
// cin >> t;
rep1(i, t) {
solve(i);
}
return 0;
}
