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 <bits/stdc++.h>
using namespace std;

#define debug(x) cout <<"DEBUG"<<" "<<#x<<":"<<x<<'\n'

//↓0-indexか1-indexでn回繰り返し
#define rep(i,n) for(int i = 0; i < ((int)(n)); i++)   //0-index昇順
#define rep1(i,n) for(int i = 1; i <= ((int)(n)); i++) //1-index昇順
#define rrep(i,n) for(int i = ((int)(n)-1); i >= 0; i--)  //0-index降順
#define rrep1(i,n) for(int i = ((int)(n)); i >= 1; i--)   //1-index降順
//rep2 -> 第二引数 m から n 回繰り返し ex) m=5 n=3 なら i=5,i=6,i=7 まで
#define rep2(i,m,n) for(int i = ((int)(m)); i < ((int)(n))+((int)(m)); i++)
#define SIZE(x) ((int)((x).size()))
#define all(x) (x).begin(),(x).end()
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<vi> vii; //2次元配列
typedef vector<ll> vll;
typedef vector<string> vs;
template<class T,class U> inline bool chmax(T &a, const U &b){if(a<b){a=b;return 1;}return 0;}
template<class T,class U> inline bool chmin(T &a, const U &b){if(a>b){a=b;return 1;}return 0;}

/* 円周率は M_PI を使おうね */
const ll infll = (1LL<<62)-1;
const int inf = (1<<30)-1;
/*GRAPH_TEMPLATE=============================================*/
template<typename T>
class Edge {
public:
    int srch; //searched?探索状態記録「null = -1」
    int to; //辺の行き先
    T cost; //辺の重み
    Edge(int t, T w) : srch(-1), to(t), cost(w) {}
    Edge(int src, int t, T w) : srch(src), to(t), cost(w) {}
};
template<typename T>
using W_Graph = vector<vector<Edge<T>>>;   //重み付きグラフ
using Graph = vector<vector<int>>;         //通常グラフ
template<typename T>
using Matrix = vector<vector<T>>;          //隣接行列(使わなさそう...)

/*FUNCs=================================================*/


/*MAIN==================================================*/
signed main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);   //cin cout 高速化
    //cout << fixed << setprecision(10);

    int n, m; cin >> n >> m;
    W_Graph<int> G(n);
    rep(i,m) {
        int l, r, c;
        cin >> l >> r >> c;
        --l; --r;
        G[l].push_back(Edge<int>(r, c));
        G[r].push_back(Edge<int>(l, c));
        for(int j = l; j < r; ++j) {
            G[j+1].push_back(Edge<int>(j, 0));  //こうすればl~r間の全ての頂点が
                                               //重みcの辺で結ばれたことになる
        }
    }

    //dijkstra
    using p = pair<int,int>;
    priority_queue<p, vector<p>, greater<p>> que;
    vi min_dist(n+10, inf);
    min_dist[0] = 0;
    que.emplace(min_dist[0], 0);
    while(que.size()) {
        int dist;
        int index;
        tie(dist, index) = que.top();
        que.pop();
        if(dist > min_dist[index]) continue;

        for(auto edge : G[index]) {
            int new_dist = dist+edge.cost;
            if(min_dist[edge.to] <= new_dist) continue;
            chmin(min_dist[edge.to], new_dist);
            que.emplace(min_dist[edge.to], edge.to);
        }
    }
    if(min_dist[n-1]!=inf) cout << min_dist[n-1] << '\n';
    else cout << -1 << '\n';
}