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: Time Limit Exceeded)

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <tuple>
#include <cstring>
#include <functional>
using namespace std;
typedef long long ll;
typedef pair<long long, long long> P;
#define rep(i, n) for(long long i=0; i<n; i++)
#define reps(i, s, e) for(long long i=s; i<e; i++)
#define repr(i, n) for(long long i=n-1; i>=0; i--)
#define reprs(i, s, e) for(long long i=e-1; i>=s; i--)

int main(){

    cin.tie(0);
    ios::sync_with_stdio(false);

    ll n, m; cin >> n >> m;
    ll l[m], r[m], c[m];
    rep(i, m){
        cin >> l[i] >> r[i] >> c[i];
        l[i]--; r[i]--;
    }

    vector<P> vec[n];
    rep(i, m){
        reps(j, l[i] + 1, r[i] + 1){
            vec[j].push_back({l[i], c[i]});
        }
    }

    bool skip[n];
    fill(skip, skip+n, true);
    rep(i, m){
        skip[l[i]] = false;
        skip[r[i]] = false;
    }


    ll dp[n];
    fill(dp, dp+n, 1e18);
    dp[0] = 0;

    reps(i, 1, n){
        if(vec[i].size() == 0){
            cout << -1 << endl;
            return 0;
        }
       if(skip[i]) continue;

        ll tempmin = 1e18;
        rep(j, vec[i].size()){
            ll start = vec[i][j].first;
            ll cost  = vec[i][j].second;

            dp[i] = min(dp[i], dp[start] + cost);
        }
    }

    cout << dp[n-1] << endl;

    return 0;
}