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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::input;
use proconio::marker::Usize1;
use std::collections::BinaryHeap;
use std::cmp::Reverse;

fn main()
{
    input! {
    n:usize,
    m:usize,
    edge:[(Usize1,Usize1,i64);m],
    }

    let mut v = vec![vec![]; n];
    for i in 1..n {
        v[i].push((i - 1, 0));
    }
    for (a, b, c) in edge {
        v[a].push((b, c));
    }

    let inf = 1i64 << 60;
    let mut dist = vec![inf; n];

    let mut q = BinaryHeap::new();
    q.push(Reverse((0, 0)));

    while let Some(Reverse((dst, idx))) = q.pop() {
        if dist[idx] < dst {
            continue;
        }
        for &(nxt, cost) in v[idx].iter() {
            let dst = dst + cost;
            if dist[nxt] <= dst {
                continue;
            }
            dist[nxt] = dst;
            q.push(Reverse((dst, nxt)));
        }
    }

    let ans = if dist[n - 1] == inf {
        -1
    } else {
        dist[n - 1]
    };
    println!("{}", ans);
}