## USACO 2019 December Contest, Platinum

## Problem 1. Greedy Pie Eaters

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Farmer John may choose a sequence of cows $c_1,c_2,\ldots, c_K,$ after which the selected cows will take turns eating in that order. Unfortunately, the cows don't know how to share! When it is cow $c_i$'s turn to eat, she will consume all of the pies that she enjoys --- that is, all remaining pies in the interval $[l_{c_i},r_{c_i}]$. Farmer John would like to avoid the awkward situation occurring when it is a cows turn to eat but all of the pies she enjoys have already been consumed. Therefore, he wants you to compute the largest possible total weight ($w_{c_1}+w_{c_2}+\ldots+w_{c_K}$) of a sequence $c_1,c_2,\ldots, c_K$ for which each cow in the sequence eats at least one pie.

#### SCORING:

- Test cases 2-5 satisfy $N\le 50$ and $M\le 20$.
- Test cases 6-9 satisfy $N\le 50.$

#### INPUT FORMAT (file pieaters.in):

The first line contains two integers $N$ and $M$ $\left(1\le M\le \frac{N(N+1)}{2}\right)$.The next $M$ lines each describe a cow in terms of the integers $w_i, l_i$, and $r_i$.

#### OUTPUT FORMAT (file pieaters.out):

Print the maximum possible total weight of a valid sequence.#### SAMPLE INPUT:

2 2 100 1 2 100 1 1

#### SAMPLE OUTPUT:

200

In this example, if cow 1 eats first, then there will be nothing left for cow 2 to eat. However, if cow 2 eats first, then cow 1 will be satisfied by eating the second pie only.

Problem credits: Benjamin Qi