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: Accepted)

#include <iostream>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <bitset>
#include <complex>
#include <functional>

// output
#define SPBR(w, n) std::cout<<(w + 1 == n ? '\n' : ' ');
#define YES cout << "YES" << endl
#define Yes cout << "Yes" << endl
#define NO cout << "NO" << endl
#define No cout << "No" << endl
// utility
#define ALL(i) (i).begin(), (i).end()
#define FOR(i, a, n) for(int i=(a);i<(n);++i)
#define RFOR(i, a, n) for(int i=(n)-1;i>=(a);--i)
#define REP(i, n) for(int i=0;i<int(n);++i)
#define RREP(i, n) for(int i=int(n)-1;i>=0;--i)
#define IN(a, x, b) (a<=x && x<b)
#define OUT(a, x, b) (x<a || b<=x)
template<class T> inline T chmax(T & a, const T b) { return a = (a < b) ? b : a; }
template<class T> inline T chmin(T& a, const T b) { return a = (a > b) ? b : a; }

// type/const
#define int ll
using ll = long long;
using ull = unsigned long long;
using ld = long double;
const int MOD = 1000000007;
/* const int MOD = 1000000007; */
const int INF = 1e18;
const double PI = acos(-1);

using namespace std;

vector<int> dijkstra(int i,int n, vector<vector<pair<int, int>>> graph) {
  vector<int> d(n, INF);
  d[i] = 0;
  priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> q;
  q.push(make_pair(0, i));
  while (!q.empty()) {
    pair<int, int> p = q.top();
    q.pop();
    int v = p.second;
    if (d[v] < p.first) {
      continue;
    }
    for (auto x : graph[v]) {
      if (d[x.first] > d[v] + x.second) {
        d[x.first] = d[v] + x.second;
        q.push(pair<int, int>(d[x.first], x.first));
      }
    }
  }
  return d;
}

signed main() {
  int N, M;
  cin >> N >> M;

  vector<vector<pair<int, int>>> graph(N);
  REP(i, M){
    int x, y, d;
    cin >> x >> y >> d;
    x--; y--;
    graph[x].emplace_back(y, d);
  }
  FOR(i, 1, N){
    graph[i].emplace_back(i-1, 0);
  }

  vector<int> dist = dijkstra(0, N, graph);
  if(dist[N-1] == INF) dist[N-1] = -1;
  cout << dist[N-1] << endl;

  return 0;
}