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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

n,m = map(int,input().split())
lrc = []
for _ in range(m):
  lrc.append(list(map(int,input().split())))
import heapq
heapq.heapify(lrc)
arrive = [[1,0]]
heapq.heapify(arrive)
#print(arrive)
while True:
  if arrive == []:
    print(-1)
    break
  point,dis = heapq.heappop(arrive)
  if point == n:
    print(dis)
    break
  if arrive != []:
    if dis >= arrive[0][1]:
      continue
  while True:
    if lrc == []:
      break
    if lrc[0][0] <= point:
      l,r,c = heapq.heappop(lrc)
      if r > point:
        heapq.heappush(arrive,[r,dis+c])
    else:
      break