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: Time Limit Exceeded)

#include <bits/stdc++.h>
     
using namespace std;

long long int INF = 1e18;
vector <pair<int,int>> adj[100001];
long long int dist[100001];

int main(void)
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	int n,m,l,r,c;

	cin >> n >> m;

	for(int i=0;i<m;i++)
	{
		cin >> l >> r >> c;
		adj[l].push_back(make_pair(r,c));
		adj[r].push_back(make_pair(l,c));
	}

	for(int i=2;i<=n;i++)
	{
		adj[i].push_back(make_pair(i-1,0));
	}

	fill(dist,dist+100001,INF);
	dist[1] = 0;
	priority_queue <pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>> pque;
	pque.push(make_pair(dist[1],1));

	while(!pque.empty())
	{
		int now = pque.top().second;
		if(dist[now] < pque.top().first)
		{
			pque.pop();
			continue;
		}
		pque.pop();

		for(auto it : adj[now])
		{
			int cost = it.second;
			int next = it.first;
			if(dist[next] > dist[now] + cost)
			{
				dist[next] = dist[now] + cost;
				pque.push(make_pair(dist[next],next));
			}
		}
	}

	if(dist[n]==INF)
	{
		cout << -1 << '\n';
	}
	else
	{
		cout << dist[n] << '\n';
	}

	return 0;
}