#!/usr/bin/env python3
n = int(input())
# 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.
dp = [[-(10**18)] * (i + 2) for i in range(n + 1)]
dp[0][0] = 0
last_e = 0
for i in range(1, n + 1):
    s, p, e, a = 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.
        # NOTE: WA because this should also add `s - last_e`.
        if a <= dp[i - 1][k - 1]:
            dp[i][k] = max(dp[i][k], dp[i - 1][k - 1] - a + e - p + s - last_e)
    last_e = e
maxk = 0
for k in range(n + 1):
    if dp[n][k] >= 0:
        maxk = k
print(maxk)