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:
import heapq
import sys

class rode():
    def __init__(self,x,y):
        self.x = x
        self.y = y
    def __lt__(self, other):
        return self.y < other.y
    def __gt__(self, other):
        return other > self
    def __le__(self, other):
        return not(other > self)
    def __ge__(self, other):
        return not(self > other)

n,m = map(int,input().split())
da = [[] for i in range(n)]
for i in range(m):
    l,r,c = map(int,input().split())
    da[l-1].append([r-1,c])
for i in range(1,n):
    da[i].append([i-1,0])
pq=[]
ju = [False for i in range(n)]
ju[0] = True
for i in da[0]:
    heapq.heappush(pq, rode(i[0],i[1]))
while(len(pq)):
    now = heapq.heappop(pq)
    if now.x == n-1:
        print(now.y)
        sys.exit()
    if ju[now.x]:
        continue
    ju[now.x] = True
    for i in da[now.x]:
        if ju[i[0]]:
            continue
        heapq.heappush(pq,rode(i[0],now.y+i[1]))
print(-1)