Hacker, pack your bags!
Contest: Codeforces Round 422 (Div. 2)
Difficulty rating: 1600
Tags: binary search, greedy, implementation, sortings

It's well known that the best way to distract from something is to do one's favourite thing. Job is such a thing for Leha.

So the hacker began to work hard in order to get rid of boredom. It means that Leha began to hack computers all over the world. For such zeal boss gave the hacker a vacation of exactly x days. You know the majority of people prefer to go somewhere for a vacation, so Leha immediately went to the travel agency. There he found out that n vouchers left. i-th voucher is characterized by three integers li, ri, costi — day of departure from Vičkopolis, day of arriving back in Vičkopolis and cost of the voucher correspondingly. The duration of the i-th voucher is a value ri - li + 1.

At the same time Leha wants to split his own vocation into two parts. Besides he wants to spend as little money as possible. Formally Leha wants to choose exactly two vouchers i and j (i ≠ j) so that they don't intersect, sum of their durations is exactly x and their total cost is as minimal as possible. Two vouchers i and j don't intersect if only at least one of the following conditions is fulfilled: ri < lj or rj < li.

Help Leha to choose the necessary vouchers!

Input format:
The first line contains two integers n and x (2 ≤ n, x ≤ 2·105) — the number of vouchers in the travel agency and the duration of Leha's vacation correspondingly.

Each of the next n lines contains three integers li, ri and costi (1 ≤ li ≤ ri ≤ 2·105, 1 ≤ costi ≤ 109) — description of the voucher.

Output format:
Print a single integer — a minimal amount of money that Leha will spend, or print  - 1 if it's impossible to choose two disjoint vouchers with the total duration exactly x.

Constraints:
Time limit: 2.0 s
Memory limit: 256.0 MB

Note:
In the first sample Leha should choose first and third vouchers. Hereupon the total duration will be equal to (3 - 1 + 1) + (6 - 5 + 1) = 5 and the total cost will be 4 + 1 = 5.

In the second sample the duration of each voucher is 3 therefore it's impossible to choose two vouchers with the total duration equal to 2.

Example input:
4 5
1 3 4
1 2 5
5 6 1
1 2 4

Example output:
5

Example input:
3 2
4 6 3
2 4 1
3 5 4

Example output:
-1

Editorial:
Let's sort all the vouchers by their left border li. Let's consider the vouchers in this sorted order. Now we want to consider i-th one. The duration of the i-th voucher is ri - li + 1. Consequentally only the vouchers with the duration x - ri + li - 1 can make a couple with the i-th one. At the same time you shouldn't forget that the value of this expression may be negative. You should check it. Besides let's find a couple for the i-th voucher among all the vouchers k for which rk < li.
To implement this solution let's keep an array bestCost. bestCost[j] denotes the minimal cost of the voucher with the duration equal to exactly j on the considering prefix (i.e. we should consider only such vouchers k with rk < li in bestCost). Thus, it's enough to consider vouchers in order of increasing of their left borders and update the array bestCost.
Total complexity is $$O(n \log n)$$.