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: Accepted)

from heapq import heappop, heappush

N,M = map(int,input().split())
LRC = [list(map(int,input().split())) for _ in range(M)]

INF = float('inf')
def dijkstra(s,n,g): # sからの最短距離 頂点数n g[v]=[(u1,w1),(u2,w2)....]:vからuへの有向辺があり、重みw
  que = []
  d = [INF for _ in range(n)]
  d[s] = 0
  heappush(que,(0,s))
  while len(que) > 0:
    p = heappop(que)
    v = p[1]
    if d[v] < p[0]:
      continue
    for u,w in g[v]:
      if d[u] > d[v]+w:
        d[u] = d[v]+w
        heappush(que,(d[u],u))
  return d

G = [[] for _ in range(N)]
for i in range(N-1):
  G[i+1].append((i,0))
for l,r,c in LRC:
  l -= 1
  r -= 1
  G[l].append((r,c))

p = dijkstra(0,N,G)[N-1]
print(p if p != INF else -1)