Problem:
Score : 500 points

Problem Statement

Takahashi is at an all-you-can-eat restaurant.

The restaurant offers N kinds of dishes. It takes A_i minutes to eat the i-th dish, whose deliciousness is B_i.

The restaurant has the following rules:

You can only order one dish at a time. The dish ordered will be immediately served and ready to eat.

You cannot order the same kind of dish more than once.

Until you finish eating the dish already served, you cannot order a new dish.

After T-0.5 minutes from the first order, you can no longer place a new order, but you can continue eating the dish already served.

Let Takahashi's happiness be the sum of the deliciousness of the dishes he eats in this restaurant.

What is the maximum possible happiness achieved by making optimal choices?

Constraints

2 \leq N \leq 3000

1 \leq T \leq 3000

1 \leq A_i \leq 3000

1 \leq B_i \leq 3000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N T
A_1 B_1
:
A_N B_N

Output

Print the maximum possible happiness Takahashi can achieve.

Sample Input 1

2 60
10 10
100 100

Sample Output 1

110

By ordering the first and second dishes in this order, Takahashi's happiness will be 110.

Note that, if we manage to order a dish in time, we can spend any amount of time to eat it.

Sample Input 2

3 60
10 10
10 20
10 30

Sample Output 2

60

Takahashi can eat all the dishes within 60 minutes.

Sample Input 3

3 60
30 10
30 20
30 30

Sample Output 3

50

By ordering the second and third dishes in this order, Takahashi's happiness will be 50.

We cannot order three dishes, in whatever order we place them.

Sample Input 4

10 100
15 23
20 18
13 17
24 12
18 29
19 27
23 21
18 20
27 15
22 25

Sample Output 4

145

Solution in PyPy3 (2.4.0) (judge verdict: Accepted)

import sys
from operator import itemgetter

input = sys.stdin.readline

#################
# 定数
#################
TIME = 0
VAL = 1

##################
# 入力処理
##################
N, T = [int(a) for a in input().split()]
time_value = [(-1, -1)] + [None] * (N)
for i in range(1, N + 1):
    time_value[i] = tuple(int(a) for a in input().split())

# Sort the array in the increasing order of value
time_value.sort(key = itemgetter(VAL))

###########################
# 動的計画法（Knapsack problem）のテーブル埋め
# 時間の最大値は T - 1
##########################
dp = [[-1] * T for _ in range(N + 1)] # (N+1)行 x T列 2次元リスト

for t in range(0, T):
    dp[0][t] = 0

for n in range(1, N + 1):
    dp[n][0] = 0
    for t in range(1, T):
        if time_value[n][TIME] > t:
            dp[n][t] = dp[n - 1][t]
        else:
            dp[n][t] = max(dp[n - 1][t],
                             time_value[n][VAL] + dp[n - 1][t - time_value[n][TIME]])

##########################
# DPテーブルを用いて問題を解く
##########################

# t = T - 1に食べ始める料理が N, N-1, N-2, ..., 2, 1それぞれの場合で
# 美味しさの最大値を出し、更にその最大値を取る

# 最後に食べる料理がNの場合
val_acum = time_value[N][VAL]
t = T - 1  # N以外の料理を食べるのに使える時間
max_val = val_acum + dp[N - 1][t]

for n in range(N - 1, 0, -1):  # 最後に食べる料理が n = N-1, N-2, ..., 2, 1の場合
    val_acum += time_value[n][VAL] # 料理 N, N-1, ..., nの美味しさ合計
    t -= time_value[n + 1][TIME] # その他の料理 1, 2, ..., n-1を食べるのに使える時間
    if t < 0:
        break
    else:
        max_val = max(max_val, val_acum + dp[n - 1][t])

print(max_val)