#include <bits/stdc++.h>
// CCO '16 P2 - Splitting Hares

#define EPS 1e-7

using namespace std;

int x[4005],y[4005],w[4005];
vector<int> l,r;
int cur;

inline double slope(int i1,int i2) {
    return ((double) (y[i1] - y[i2])) / (x[i1] - x[i2]);
}

bool cmp(int i1,int i2) {
    return slope(i1,cur) < slope(i2,cur);
}

inline bool eq(double d1,double d2) {
    return (abs(d1 - d2) < EPS);
}

int main() {
    int N,i,j,k,sum1,sum2,ans,dec1,dec2,temp;
    double nxt;

    scanf("%d",&N);

    for (i = 0;i < N;i++)
        scanf("%d%d%d",&x[i],&y[i],&w[i]);

    ans = 0;
    for (i = 0;i < N;i++) {
        sum1 = 0;
        sum2 = 0;
        dec1 = 0;
        dec2 = 0;
        l.clear();
        r.clear();

        for (j = 0;j < N;j++) {
            if (x[j] == x[i]) {
                sum1 += w[j];
                sum2 += w[j];

                if (y[j] > y[i])
                    dec1 += w[j];
                else if (y[j] < y[i])
                    dec2 += w[j];
            }
            else if (x[j] > x[i]) {
                r.push_back(j);
                sum2 += w[j];
            }
            else {
                l.push_back(j);
                sum1 += w[j];
            }
        }

        cur = i;
        sort(l.begin(),l.end(),cmp);
        sort(r.begin(),r.end(),cmp);

        j = 0;
        k = 0;
        while (j < l.size() || k < r.size()) {
            temp = max(max(dec1,dec2) + max(w[i],0),dec1 + dec2 + w[i]);
            ans = max(ans,max(temp,0) + max(sum1,sum2) - dec1 - dec2 - w[i]);

            if (j == l.size())
                nxt = slope(r[k],i);
            else if (k == r.size())
                nxt = slope(l[j],i);
            else
                nxt = min(slope(l[j],i),slope(r[k],i));

            sum1 -= dec1;
            sum2 -= dec2;
            dec1 = 0;
            dec2 = 0;

            while (j < l.size() && eq(slope(l[j],i),nxt)) {
                sum2 += w[l[j]];
                dec1 += w[l[j]];
                j++;
            }

            while (k < r.size() && eq(slope(r[k],i),nxt)) {
                sum1 += w[r[k]];
                dec2 += w[r[k]];
                k++;
            }
        }
    }

    printf("%d\n",ans);

    return 0;
}