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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(){

  int n, k, pos = 0, cont = 0, boole = 0;

  scanf("%d%d", &n, &k);

  int vet[n][3];

  for(int i = 0; i < n; i++) {
    scanf("%d%d%d", &vet[i][0], &vet[i][1], &vet[i][2]);
    if (vet[i][0] == 1){
      if (vet[i][1] > pos && vet[i][1] != n){
        pos = vet[i][1];
        cont = vet[i][2];
      }
    }
  }

  for (int i = 0; i < n; i++){
    if (vet[i][0] <= pos && vet[i][1] == k){
      cont += vet[i][2];
      boole= 1;
      break;
  }
}

  for (int i = 0; i < n; i ++){
    if (vet[i][0] == 1 && vet[i][1] == k){
      if (vet[i][2] < cont){
        printf("%d\n", vet[i][2]);
        break;
      }
    }
  }


  if (!boole)
    printf("-1\n");
  else
    printf("%d\n", cont);




  return 0;
}