#include <iostream>
using namespace std;
// DP[i][j] = maximum free time available up to s_i when having studied for (j <= i) of exam 0, 1, ..., (i-1).
int n, result, a[2000], time_passing[2000], time_failing[2000], DP[2001][2001];

int main() {
	cin >> n; // Assume s[i] = p[i].
	for (int i = 0, si; i < n; i++) {
		cin >> si >> time_passing[i] >> time_failing[i] >> a[i];
		if (i) time_passing[i-1] = si - time_passing[i-1], time_failing[i-1] = si - time_failing[i-1];
		else DP[0][0] = si;
	}
	for (int i = 0; i < n; i++) {
		for (int j = 0; j <= i; j++) {
			DP[i + 1][j] = DP[i][j] < 0 ? -1 : DP[i][j] + time_failing[i]; // Do not study for exam i:
		}
		DP[i + 1][i + 1] = -1;

		for (int j = 0; j <= i + 1; j++) {
			// Doe wat stoms: kijk of ik examen k-1 kan halen, als ik `i` nu doe.
			int extra_time = 0;
			for (int k = i + 1; k <= n; k++) {
				if (DP[i][j] + extra_time >= a[i])
					DP[k][j + 1] = max(DP[k][j + 1], DP[i][j] + extra_time - a[k-1] + time_passing[k-1]);
				extra_time += time_failing[k-1];
			}
		}
	}

/*
	for (int i = 0; i <= n; i++) {
		for (int j = 0; j <= i; j++)
			cerr << DP[i][j] << " ";
		cerr << endl;
	}
*/

	for (result = n; result > 0 && DP[n][result] < 0; result--);
	cout << result << endl;
}