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)

#pragma GCC optimize("O3")
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <fstream>
#include <functional>
#include <bitset>
using namespace std;

using ll = long long;
using P = pair<int, int>;
using T = tuple<ll, ll, ll>;

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;}

constexpr int MOD = 1e9 + 7;
constexpr int inf = 1e9;
constexpr long long INF = 1e18;
constexpr double pi = acos(-1);
constexpr double EPS = 1e-10;

int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};

struct SegmentTree{
private:
    int n;
    vector<ll> node;

public:
    // Range Minumum Query
    SegmentTree(vector<ll> v){
        int sz = v.size();
        n = 1; while(n < sz) n *= 2;
        node.resize(2*n-1, 0);
        for(int i=0; i<sz; i++) node[i+n-1] = v[i];
        for(int i=n-2; i>=0; i--) node[i] = min(node[2*i+1], node[2*i+2]);
    }

    void update(int x, int val){
        x += (n - 1);
        node[x] = val;
        while(0 < x){
            x = (x - 1) / 2;
            node[x] = min(node[2*x+1], node[2*x+2]);
        }
    }

    ll getmin(int a, int b, int k=0, int l=0, int r=-1){
        if(r < 0) r = n;
        if(r <= a || b <= l) return INF;
        if(a <= l && r <= b) return node[k];

        ll vl = getmin(a, b, 2*k+1, l, (l+r)/2);
        ll vr = getmin(a, b, 2*k+2, (l+r)/2, r);
        
        return min(vl, vr);
    }
};

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n, m; cin>>n>>m;
    vector<ll> dist(n, INF);
    dist[0] = 0; 
    SegmentTree seg(dist);
    vector<ll> dist2(n, INF);
    dist2[0] = 0;
    vector<T> v;
    for(int i=0; i<m; i++){
        int l, r, c; cin>>l>>r>>c;
        v.emplace_back(l, r, c);
    }
    sort(v.begin(), v.end());

    for(int i=0; i<m; i++){
        ll l, r, c;
        tie(l, r, c) = v[i];
        //cout << l << " " << r << " " << c << endl;
        l--, r--;
        ll val = seg.getmin(l, r + 1);
        val += c;
        if(dist2[r] > val){
            seg.update(r, val);
        }
        chmin(dist2[r], val);
    }

    cout << (dist2[n-1] == INF ? -1 : dist2[n-1]) << endl;
    return 0;
}