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 l_i, r_i, cost_i — day of departure from Vi?kopolis, day of arriving back in Vi?kopolis and cost of the voucher correspondingly. The duration of the i-th voucher is a value r_i?-?l_i?+?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: r_i?<?l_j or r_j?<?l_i.

Help Leha to choose the necessary vouchers!

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 l_i, r_i and cost_i (1?≤?l_i?≤?r_i?≤?2·10⁵,?1?≤?cost_i?≤?10⁹) — description of the voucher.

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.

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.

从n个里找出两个来，使他们的值加起来等于x并且不在一个区间里。

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <vector>
 4 using namespace std;
 5 const int N = 2e5+10;
 6 vector<pair<int,int> > a[N], b[N];
 7 typedef pair<int,int> P;
 8 const int INF = 2e9+10;
 9 int c[N];
10 int main() {
11     int n, k, l, r, cc;
12     cin >> n >> k;
13     for(int i = 0; i < n; i ++) {
14         cin >> l >> r >> cc;
15         a[l].push_back(make_pair(r-l+1, cc));
16         b[r].push_back(make_pair(r-l+1, cc));
17     }
18     for(int i = 0; i < N; i ++) c[i] = INF;
19     int ans = INF;
20     for(int i = 0; i < N; i ++) {
21         for(int j = 0; j < a[i].size(); j ++) {
22             P p = a[i][j];
23             if(p.first>=k) continue;
24             else if(c[k-p.first]<INF) ans=min(ans,p.second+c[k-p.first]);
25         }
26         for(int j = 0; j < b[i].size(); j ++) {
27             P p = b[i][j];
28             c[p.first]=min(c[p.first],p.second);
29         }
30     }
31     printf("%d\n",ans==INF?-1:ans);
32     return 0;
33 }