CF422 div2 C. Hacker, pack your bags!

C. Hacker, pack your bags!

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

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

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.

Examples

input

4 51 3 41 2 55 6 11 2 4

output

5

input

3 24 6 32 4 13 5 4

output

-1

不难想到哈希，这里需要采用区间左右端点的哈希。

一个数组记录当前存在的天数为v的一个最小花费，初始为无穷大。

对于当前点，先看是左端点的，如果存在x-v的一个最小花费区间，那么更新答案。

然后看右端点，直接更新天数为v的一个最小花费。

这样就能够避免一后效性和重复了。

#include<bits/stdc++.h>using namespace std;const int MAXN=2e5+7;const int inf = 2e9+1;int n,m;struct node{    int l,r;    int cost;    int v;    int flag;    bool operator < (const node &a)const    {        return flag < a.flag;    }} p[MAXN];vector<node>ha[MAXN];int vis[MAXN];int main(){    scanf("%d%d",&n,&m);    for(int i = 0 ; i < n ; ++i)    {        scanf("%d%d%d",&p[i].l,&p[i].r,&p[i].cost);        p[i].v = p[i].r - p[i].l + 1;        p[i].flag = 1;        ha[p[i].l].push_back(p[i]);        p[i].flag = 2;        ha[p[i].r].push_back(p[i]);    }    for(int i = 1 ; i <= 200000; ++i)    {        //sort(ha[i].begin(),ha[i].end());        vis[i] = inf;    }    int MIN = inf;    for(int i = 1 ; i <= 200000 ; ++i)    {        for(int j = 0 , l = ha[i].size() ; j < l ;++j)        {            int v = ha[i][j].v;            int cost = ha[i][j].cost;            int flag = ha[i][j].flag;            if(flag == 1)            {                if(v < m && vis[m-v] != inf)                {                    MIN = min(MIN,cost + vis[m-v]);                }            }            else vis[v] = min(vis[v],cost);        }    }    if(MIN == inf)puts("-1");    else printf("%d\n",MIN);}