Problem:
Score : 600 points

Problem Statement

We have N points numbered 1 to N arranged in a line in this order.

Takahashi decides to make an undirected graph, using these points as the vertices.
In the beginning, the graph has no edge. Takahashi will do M operations to add edges in this graph.
The i-th operation is as follows:

The operation uses integers L_i and R_i between 1 and N (inclusive), and a positive integer C_i. For every pair of integers (s, t) such that L_i \leq s < t \leq R_i, add an edge of length C_i between Vertex s and Vertex t.

The integers L_1, ..., L_M, R_1, ..., R_M, C_1, ..., C_M are all given as input.

Takahashi wants to solve the shortest path problem in the final graph obtained. Find the length of the shortest path from Vertex 1 to Vertex N in the final graph.

Constraints

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq L_i < R_i \leq N

1 \leq C_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N M
L_1 R_1 C_1
:
L_M R_M C_M

Output

Print the length of the shortest path from Vertex 1 to Vertex N in the final graph.
If there is no shortest path, print -1 instead.

Sample Input 1

4 3
1 3 2
2 4 3
1 4 6

Sample Output 1

5

We have an edge of length 2 between Vertex 1 and Vertex 2, and an edge of length 3 between Vertex 2 and Vertex 4, so there is a path of length 5 between Vertex 1 and Vertex 4.

Sample Input 2

4 2
1 2 1
3 4 2

Sample Output 2

-1

Sample Input 3

10 7
1 5 18
3 4 8
1 3 5
4 7 10
5 9 8
6 10 5
8 10 3

Sample Output 3

28

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
#include <time.h>
#include <string.h>
#include <vector>
#include <queue>
#include <map>
using namespace std;
struct EDGE {
	int x, y, z;
	inline int operator<(const EDGE &e2) const {
		return x < e2.x;
	};
} e[100005];
int n, m;
long long v[400005], lazy[400005];
void pushd(int nown) {
	lazy[nown << 1]= min(lazy[nown << 1], lazy[nown]);
	lazy[nown << 1 | 1]= min(lazy[nown << 1 | 1], lazy[nown]);
	v[nown << 1]= min(v[nown << 1], lazy[nown]);
	v[nown << 1 | 1]= min(v[nown << 1 | 1], lazy[nown]);
	lazy[nown]= 0x3f3f3f3f3f3f3f3f;
	return;
}
void update(int nown, int l, int r, int ml, int mr, long long c) {
	if(ml <= l && r <= mr) {
		lazy[nown]= min(lazy[nown], c);
		v[nown]= min(v[nown], c);
		return;
	}
	if(lazy[nown] != 0x3f3f3f3f3f3f3f3f) pushd(nown);
	int mid= (l + r) >> 1;
	if(mid >= ml) update(nown << 1, l, mid, ml, mr, c);
	if(mid < mr) update(nown << 1 | 1, mid + 1, r, ml, mr, c);
	return;
}
long long query(int nown, int l, int r, int mlr) {
	if(l == r) return v[nown];
	if(lazy[nown] != 0x3f3f3f3f3f3f3f3f) pushd(nown);
	int mid= (l + r) >> 1;
	if(mid >= mlr) return query(nown << 1, l, mid, mlr);
	return query(nown << 1 | 1, mid + 1, r, mlr);
}
int main() {
	for(int i= 1; i <= 400000; i++) v[i]= lazy[i]= 0x3f3f3f3f3f3f3f3f;
	cin >> n >> m;
	for(int i= 1; i <= m; i++) {
		cin >> e[i].x >> e[i].y >> e[i].z;
	}
	update(1, 1, n, 1, 1, 0);
	sort(e + 1, e + m + 1);
	for(int i= 1; i <= m; i++) {
		update(1, 1, n, e[i].x, e[i].y, query(1, 1, n, e[i].x) + e[i].z);
	}
	int res= query(1, 1, n, n);
	if(res == 0x3f3f3f3f3f3f3f3f) res= -1;
	cout << res << endl;
	return 0;
}