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 Ruby (2.3.3) (judge verdict: Wrong Answer)

N,M=gets.chomp.split(" ").map(&:to_i)
edge_list = []
M.times do
    l,r,c=gets.chomp.split(" ").map(&:to_i)
    (l..(r-1)).each do |ll|
        ((ll+1)..r).each do |rr|
            edge_list.push([ll, rr, c])
            edge_list.push([rr, ll, c])
        end
    end
end

INF = 10**10
def bellman_ford(edge_list, s, max_count)
    d = Array.new(N+1, INF)
    d[s] = 0
    c = 0
    loop do
      is_update = false
      edge_list.each do |(u, v, c)|
        if d[u] != INF && d[u] + c < d[v]
          d[v] = d[u] + c
          is_update = true
        end
      end
      return d if !is_update
      c += 1
      return nil if c == max_count
    end
end

ret = bellman_ford(edge_list, 1, N)

if ret == nil || ret[N] == INF
    puts "-1"
else
    puts ret[N]
end