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: Accepted)

#include<stdio.h>
long long CO[200010];
long long R=1,C=1,H[2000010],N[2000010];
//評価関数（いまはMIN）
int hyouka(int a,int b){
  if(C<b)return 1;
  if(C<a||b==0)return 0;
  return N[H[a]]<N[H[b]]?1:0;
}
//挿入関数
void hin(long long a){
  int i=C++;
  for(N[H[0]=R]=a;hyouka(0,i/2);i/=2)H[i]=H[i/2];
  H[i]=R++;
}
//取り出す関数
int hout(){
  int rt=H[1],i,j=2,k=H[--C];
  for(i=1;hyouka(i,C);i=j)H[i]=H[j=i*2+1-hyouka(i*2,i*2+1)];
  H[j/2]=k;
  return rt;
}
void dijk(int v,int e,int *fr,int *to,int *co,int mi){
  int i,id[200010],ta[200010],nt[200010],f[200010]={0};
  for(i=0;i<v;i++)ta[i]=-1;
  for(i=0;i<v;i++)CO[i]=1e18+10;
  for(i=CO[mi]=0;i<e;i++){
    nt[i]=ta[fr[i]];
    ta[fr[i]]=i;
  }
  while(f[mi]-1){
    f[mi]=1;
    for(i=ta[mi];i+1;i=nt[i]){
      if(CO[to[i]]>CO[mi]+co[i])hin(CO[id[R]=to[i]]=CO[mi]+co[i]);
    }
    while(f[mi]&&C-1)mi=id[hout()];
  }
}
int main(){
  int n,m,i,r=0,a[200010],b[200010],c[200010];
  scanf("%d %d",&n,&m);
  for(i=1;i<n;i++){
    a[r]=i+1;
    b[r]=i;
    c[r++]=0;
  }
  for(i=0;i<m;i++){
    scanf("%d %d %d",&a[r],&b[r],&c[r]);
    r++;
  }
  dijk(n+1,r,a,b,c,1);
  printf("%lld\n",CO[n]>1e17?-1:CO[n]);
  return 0;
}