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 Python (3.4.3) (judge verdict: Accepted)

import sys
import heapq
sys.setrecursionlimit(10**7)
n, m = [int(i) for i in sys.stdin.readline().split()]
INF = 10**9
graph = {j:[(j-1, 1e-10)] if j > 0 else [] for j in range(n)}
for i in range(m):
    l, r, c = [int(i) for i in sys.stdin.readline().split()]
    l -= 1
    r -= 1
    graph[l].append((r, c))
    graph[r].append((l, c))
que = [(0,0)]
already = set()
res = [-1 for i in range(n)]
while len(que) > 0:
    cost, cand = heapq.heappop(que)
    if cand in already:
        continue
    res[cand] = cost
    already.add(cand)
    for _next, _next_cost in graph[cand]:
        if _next in already:
            continue
        heapq.heappush(que, (cost + _next_cost, _next))
print(int(res[-1]))