#!/usr/bin/env python3

# dp[i][k]:
# the maximal amount of leftover studying time up to e_i (the end of exam i), given that we pass k out of i exams.
(n := int(input())), (dp := [[-(10 ** 18)] * (i + 2) for i in range(n + 1)])
dp[0][0], e = 0, 0
for i in range(1, n + 1):
    last_e, (s, p, e, a) = e, map(int, input().split())
    # We can pass 0..=i of the first i exams
    for k in range(0, i + 1):
        # Skip the current exam.
        dp[i][k] = max(dp[i][k], dp[i - 1][k] + s - last_e)
        # Pass the current exam.
        if a <= dp[i - 1][k - 1] + s - last_e: dp[i][k] = max(dp[i][k], dp[i - 1][k - 1] - a + e - p + s - last_e)
print(max(k for k in range(n + 1) if dp[n][k] >= 0))