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>
// #include "../../cxx-prettyprint/prettyprint.hpp"
 
#define REP(i, n) for(int i = 0; i < (int)(n); i++)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define SORT(v, n) sort(v, v+n)
#define pb(a) push_back(a)
#define vi vector<int>
#define ddvi vector<vector<int>>
 
using namespace std;
typedef long long ll;
// typedef pair<int, int> P;
 
const ll INF = 1e18;
 
struct edge {
    int to;
    int cost;
};
 
// <最短距離, 頂点の番号>
using P = pair<int, int>;
 
int V;
vector<edge> G[100001];
vector<ll> d(100001);

void dijkstra(int s) {
    priority_queue<P, vector<P>, greater<P> > que;
    fill(d.begin(), d.end(), INF);
    d[s] = 0;
    que.push(P(0, s));
 
    while (!que.empty()) {
        P p = que.top();
        que.pop();
        int v = p.second;
        if (d[v] < p.first) continue;
 
        for (int i=0; i<G[v].size(); ++i) {
            edge e = G[v][i];
            if (d[e.to] > d[v] + e.cost) {
                d[e.to] = d[v] + e.cost;
                que.push(P(d[e.to], e.to));
            }
        }
    }
}
 
int main(){
cin.tie(0);
ios::sync_with_stdio(false);
// ここに処理を記述
 
	int N, M;
	cin >> N >> M;
 
	V = N+1;
 
	vector<int> L,R,C;
	int l,r,c;
	for(int i=0; i<M; i++){
		cin >> l >> r >> c;
		L.push_back(l);
		R.push_back(r);
		C.push_back(c);
	}
 
	for(int i=N;i>1;i--){
		edge e = {i-1, 0};
		G[i].push_back(e);
	}
	REP(i,M){
		edge e = {R[i], C[i]};
		G[L[i]].push_back(e);
	}
 
	dijkstra(1);
 
	// cout << d << "\n";
 
	cout << setprecision(15);
	
	if(d[N] == INF){
		cout << -1 << "\n";
	}
	else{
		cout << d[N] << "\n";
	}
 
return 0;
}