Problem: 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.

Examples

Input

4 3
1 3 2
2 4 3
1 4 6


Output

5


Input

4 2
1 2 1
3 4 2


Output

-1


Input

10 7
1 5 18
3 4 8
1 3 5
4 7 10
5 9 8
6 10 5
8 10 3


Output

28

Solution (in PYTHON3):

Solution:
from heapq import heappop, heappush, heapify
import sys
sys.setrecursionlimit(10**9)
input = lambda: sys.stdin.readline().rstrip()
inpl = lambda: list(map(int,input().split()))

N, M = inpl()
q = []
LRC = sorted([inpl() for _ in range(M)])
points = sorted(list({lrc[i] for i in range(2) for lrc in LRC}))
if points[0] != 1 or points[-1] != N:
    print(-1)
    exit()
positions = dict()
P = 0
prev = -1
for p in points:
    if p > prev:
        prev = p
        positions[p] = P
        P += 1
m = 0
dq = [(0,0)]
for i in range(P):
    while dq and dq[0][1] < i:
        heappop(dq)
    if not dq:
        print(-1)
        break
    d = dq[0][0]
    while m < M and positions[LRC[m][0]] == i:
        heappush(dq, (d+LRC[m][2], positions[LRC[m][1]]))
        m += 1
else:
    print(d)