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 <bits/stdc++.h>

int64_t solve();

int main()
{
	printf("%lld\n", solve());

	return 0;
}

int64_t solve()
{
	int N, M;
	scanf("%d%d", &N, &M);
	std::vector<std::array<int, 3>> edges(M);
	for (auto& e: edges)
	{
		for (auto& f: e)
			scanf("%d", &f);
		e[0]--;
	}
	std::sort(edges.begin(), edges.end());
	edges.push_back({1 << 30, 1 << 30, 1 << 30});
	using pii = std::pair<int64_t, int>;
	std::priority_queue<pii, std::vector<pii>, std::greater<pii>> pq;	// {dist, r}, [l, r)
	pq.push({0, 1});
	for (int i{}, e_i{}; i < N - 1; i++)
	{
		while (!pq.empty() && pq.top().second == i) pq.pop();
		if (pq.empty()) return -1;
		auto now{pq.top()};
		while (edges[e_i][0] == i)
		{
			pq.push({now.first + edges[e_i][2], edges[e_i][1]});
			e_i++;
		}
	}
	while (!pq.empty() && pq.top().second == N - 1)
		pq.pop();

	if (pq.empty()) return -1;
	else return pq.top().first;
}