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
n, m = [int(i) for i in input().split()]
lrc = []
for _ in range(m):
    l, r, c = [int(i) for i in input().split()]
    lrc.append([l-1, r-1, c])
lrc.sort()
a = []
j = 0
for i in range(n-1):
    new = []
    while j < m and lrc[j][0] == i:
        new.append(lrc[j])
        j += 1
    if i == 0:
        now = 0
    else:
        while True:
            if len(a) == 0:
                print(-1)
                exit()
            now = heapq.heappop(a)
            if now[1] >= i:
                heapq.heappush(a, now)
                break
        now = now[0]
    for k in new:
        heapq.heappush(a,[now + k[2], k[1]])


while True:
    if len(a) == 0:
        print(-1)
        exit()
    now = heapq.heappop(a)
    if now[1] >= n-1:
        heapq.heappush(a, now)
        break
now = now[0]
print(now)