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 Vi?kopolis, day of arriving back in Vi?kopolis and cost of the voucher correspondingly. The duration of thei-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
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) = 5and 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.
题目大意 给定n个区间，每个区间有个费用，选择两个不相交(端点也不能重叠)的区间使得它们的长度总和恰好为x，并且使得它们的费用和最小。如果无解输出-1。
根据常用套路，按照区间的长度进行分组。然后按照端点的大小(反正按左端点还是右端点都是一样的)进行排序。再拿两个数组跑一道前缀最小值和后缀最小值。
现在枚举每一条旅行方案，在和它的长度之和恰好为x的另一个数组中，可以用lower_bound和upper_bound快速找出两侧合法的的最后一个和第一个，然后更新一下就好了(详细可以看代码)。
Code
1 /**
2 * Codeforces
3 * Problem#822C
4 * Accepted
5 * Time:124ms
6 * Memory:16416k
7 */
8 #include <iostream>
9 #include <cstdio>
10 #include <ctime>
11 #include <cmath>
12 #include <cctype>
13 #include <cstring>
14 #include <cstdlib>
15 #include <fstream>
16 #include <sstream>
17 #include <algorithm>
18 #include <map>
19 #include <set>
20 #include <stack>
21 #include <queue>
22 #include <vector>
23 #include <stack>
24 #ifndef WIN32
25 #define Auto "%lld"
26 #else
27 #define Auto "%I64d"
28 #endif
29 using namespace std;
30 typedef bool boolean;
31 const signed int inf = (signed)((1u << 31) - 1);
32 const double eps = 1e-6;
33 const int binary_limit = 128;
34 #define smin(a, b) a = min(a, b)
35 #define smax(a, b) a = max(a, b)
36 #define max3(a, b, c) max(a, max(b, c))
37 #define min3(a, b, c) min(a, min(b, c))
38 template<typename T>
39 inline boolean readInteger(T& u){
40 char x;
41 int aFlag = 1;
42 while(!isdigit((x = getchar())) && x != ‘-‘ && x != -1);
43 if(x == -1) {
44 ungetc(x, stdin);
45 return false;
46 }
47 if(x == ‘-‘){
48 x = getchar();
49 aFlag = -1;
50 }
51 for(u = x - ‘0‘; isdigit((x = getchar())); u = (u << 1) + (u << 3) + x - ‘0‘);
52 ungetc(x, stdin);
53 u *= aFlag;
54 return true;
55 }
56
57 typedef class Trip {
58 public:
59 int l;
60 int r;
61 int fee;
62
63 Trip(int l = 0, int r = 0, int fee = 0):l(l), r(r), fee(fee) { }
64
65 boolean operator < (Trip b) const {
66 return l < b.l;
67 }
68 }Trip;
69
70 boolean operator < (const int& b, const Trip& a) {
71 return b < a.l;
72 }
73
74 int n, x;
75 vector<Trip>* a;
76 vector<int>* minu; // in order
77 vector<int>* mind; // in opposite order
78
79 inline void init() {
80 readInteger(n);
81 readInteger(x);
82 a = new vector<Trip>[x];
83 minu = new vector<int>[x];
84 mind = new vector<int>[x];
85 for(int i = 1, l, r, c; i <= n; i++) {
86 readInteger(l);
87 readInteger(r);
88 readInteger(c);
89 int len = r - l + 1;
90 if(len >= x) continue;
91 a[len].push_back(Trip(l, r, c));
92 }
93 }
94
95 int res = inf;
96 inline void solve() {
97 for(int i = 1; i < x; i++)
98 if(!a[i].empty()) {
99 sort(a[i].begin(), a[i].end());
100 minu[i].resize(a[i].size());
101 mind[i].resize(a[i].size());
102 for(int j = 0; j < (signed)a[i].size(); j++) {
103 if(j == 0) minu[i][j] = a[i][j].fee;
104 else minu[i][j] = min(a[i][j].fee, minu[i][j - 1]);
105 }
106 for(int j = (signed)a[i].size() - 1; j >= 0; j--) {
107 if(j == a[i].size() - 1) mind[i][j] = a[i][j].fee;
108 else mind[i][j] = min(a[i][j].fee, mind[i][j + 1]);
109 }
110 }
111
112 for(int i = 1; i < x; i++) {
113 // cout << i << ":" << endl;
114 if(!a[i].empty()) {
115 for(int j = 0; j < a[i].size(); j++) {
116 // cout << a[i][j].l << " " << a[i][j].r << endl;
117 Trip &t = a[i][j];
118 int len = t.r - t.l + 1;
119 int ulen = x - len;
120 int pos = upper_bound(a[ulen].begin(), a[ulen].end(), t.r) - a[ulen].begin();
121 if(pos != a[ulen].size()) smin(res, t.fee + mind[ulen][pos]);
122 pos = lower_bound(a[ulen].begin(), a[ulen].end(), t.l - ulen + 1) - a[ulen].begin() - 1;
123 if(pos != -1) smin(res, t.fee + minu[ulen][pos]);
124 }
125 }
126 }
127
128 if(res == inf)
129 puts("-1");
130 else
131 printf("%d\n", res);
132 }
133
134 int main() {
135 init();
136 solve();
137 return 0;
138 }