## USACO 2023 December Contest, Silver

## Problem 1. Bovine Acrobatics

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Farmer John has decided to make his cows do some acrobatics! First, FJ weighs his cows and finds that they have $N$ ($1\le N\le 2\cdot 10^5$) distinct weights. In particular, for each $i\in [1,N]$, $a_i$ of his cows have a weight of $w_i$ ($1\le a_i\le 10^9, 1\le w_i\le 10^9$).

His most popular stunt involves the cows forming *balanced towers*. A
*tower* is a sequence of cows where each cow is stacked on top of the next.
A tower is *balanced* if every cow with a cow directly above it has weight
at least $K$ ($1\le K\le 10^9$) greater than the weight of the cow directly
above it. Any cow can be part of at most one balanced tower.

If FJ wants to create at most $M$ ($1 \le M \le 10^9$) balanced towers of cows, at most how many cows can be part of some tower?

#### INPUT FORMAT (input arrives from the terminal / stdin):

The first line contains three space-separated integers, $N$, $M$, and $K$.The next $N$ lines contain two space-separated integers, $w_{i}$ and $a_i$. It is guaranteed that all $w_i$ are distinct.

#### OUTPUT FORMAT (print output to the terminal / stdout):

Output the maximum number of cows in balanced towers if FJ helps the cows form towers optimally.#### SAMPLE INPUT:

3 5 2 9 4 7 6 5 5

#### SAMPLE OUTPUT:

14

FJ can create four balanced towers with cows of weights 5, 7, and 9, and one balanced tower with cows of weights 5 and 7.

#### SAMPLE INPUT:

3 5 3 5 5 7 6 9 4

#### SAMPLE OUTPUT:

9

FJ can create four balanced towers with cows of weights 5 and 9, and one balanced tower with a cow of weight 7. Alternatively, he can create four balanced towers with cows of weights 5 and 9, and one balanced tower with a cow of weight 5.

#### SCORING:

- In inputs 3-5, $M \leq 5000$ and the total number of cows does not exceed $5000$.
- In inputs 6-11, the total number of cows does not exceed $2\cdot 10^5$.
- Inputs 12-17 have no additional constraints.

Problem credits: Eric Hsu