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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<queue>
using namespace std;

#define LL long long

const int CN = 1e5+5;
const LL INF = 0x3f3f3f3f3f3f3f2f;

LL read(){
    LL s = 0,ne = 1; char c = getchar();
    for(;c<'0'||c>'9';c=getchar()) if(c=='-') ne=-1;
    for(;c>='0'&&c<='9';c=getchar()) s=(s<<1)+(s<<3)+c-'0';
    return s*ne;
}

class fs{
  public: int to,nxt; LL di; void init(int t,int n,LL d) {to=t;nxt=n;di=d;}
}E[CN * 51];
int hd[CN],ecnt = 0;
void add(int x,int y,LL z) {E[++ecnt].init(y, hd[x], z); hd[x] = ecnt;}

int n,m;

/* DJ */
class DJ{
  public: int id; LL v; bool operator < (const DJ& a)const {return v > a.v;}
};
priority_queue<DJ> Q; 
LL d[CN]; bool vis[CN];
LL SP(int st,int ed){
    memset(vis, 0, sizeof(vis));
    memset(d, 0x3f, sizeof(d));
    Q.push((DJ){st, d[st] = 0});
    while(!Q.empty()){
        int u = Q.top().id; Q.pop();
        if(vis[u]) continue; vis[u] = true;
        for(int k = hd[u]; k; k = E[k].nxt){
            int v = E[k].to;
            if(d[v] > d[u] + E[k].di){
                d[v] = d[u] + E[k].di;
                if(!vis[u]) Q.push((DJ){v, d[v]});
            }
        }
    }
    return d[ed] < INF ? d[ed] : -1;
}

int main()
{
    n = read(); m = read();
    while(m--){
        int u = read(),v = read(); LL c = read();
        add(u, v, c); add(v, u, c);
    }
    for(int i=1;i<n;i++) add(i + 1, i, 0);

    printf("%lld", SP(1, n));
}