// This runs in linear time in O(T) where t is the maximum starting time
#include <iostream>
#include <iomanip>
using namespace std;

#define MAX_N 10000
#define MAX_T 1000000

struct hearing {
    int s,a,b;
};

hearing sched[MAX_N];


double opt[2+MAX_T];     // last meaningful start time is MAX_T; anything beyond that is exp=0
double sum[2+MAX_T];     // sum[k] = sum of opt[k]..opt[MAX_T]


int main(int argc, const char * argv[]) {
    int n;
    cin >> n;

    for (int j=0; j < n; j++) {
        int s,a,b;
        cin >> sched[j].s >> sched[j].a >> sched[j].b;
    }

    int h = n-1;

    sum[MAX_T+1] = opt[MAX_T+1] = 0;
    for (int t=MAX_T; t >= 1; t--) {
        // first compute opt[t]
        opt[t] = opt[t+1];         // could always just wait
        while (h >= 0 && sched[h].s == t) {
            int a(sched[h].a), b(sched[h].b);
            double rest = (sum[min(t+a,MAX_T+1)] - sum[min(t+b+1,MAX_T+1)])/(b-a+1);
            double possible = 1 + rest;
            opt[t] = max(opt[t], possible);
            if (argc > 1) {  // DEBUG MODE
                cerr << "If we choose to go to hearing " << h << " opt from there is " << possible << endl;
                cerr << "Opt at start time of " << h << " is " << opt[t] << endl;
            }
            h--;
        }

        // now compute sum[t]
        sum[t] = opt[t] + sum[t+1];
    }

    cout << fixed << setprecision(9) << opt[1] << endl;
}