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;
// type
typedef long long ll;
typedef long double ld;
// loop
//#define For(i, l, r, x)     for (int i = l; i < r; i+=x)
//#define ForE(i, l, r, x)    for (int i = l; i <= r; i+=x)
//#define Ford(i, r, l)       for (int i = r; i > l; i--)
//#define FordE(i, r, l)      for (int i = r; i >= l; i--)
//#define Fora(i, a)          for (auto i : a)
// I/O 
#define FAST_IO             std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
//#define PrintV(a)           Fora(ii, a) cout << ii << ' '; cout << endl;
//#define PrintVl(a)          Fora(ii, a) cout << ii << endl;
//#define PrintA(a, l, r)     for (int ii = l; ii <= r; ii++) cout << a[ii] << ' '; cout << endl;
//#define PrintAl(a, l, r)    for (int ii = l; ii <= r; ii++) cout << a[ii] << endl;
//#define Ptest(x)            return cout << x, 0;
#define setpre(n)           fixed << setprecision(n)
// pair
#define F                   first
#define S                   second
#define pii                 pair<int, int>
#define pll                 pair<ll, ll>
#define pdd                 pair<ld, ld>
// vector & !!?(string)
#define eb                  emplace_back
#define pb                  push_back
#define all(a)              a.begin(), a.end()
#define rall(a)             a.rbegin(), a.rend()
#define sz(a)               a.size()
#define len(a)              a.length()
// geometry calc    
#define pi                  acos(-1.0)
#define g_sin(a)            sin(a*pi/180)
#define g_cos(a)            cos(a*pi/180)
#define g_tan(a)            tan(a*pi/180)
// set val
#define ms0(a)              memset(a,        0, sizeof(a));
#define ms1(a)              memset(a,        1, sizeof(a));
#define msn1(a)             memset(a,       -1, sizeof(a));
#define msinf(a)            memset(a, 0x3f3f3f, sizeof(a));
// constant
const int mod1 = 998244353, mod2 = 1e9+7;
const int MAX_N = 100005, MAX_M = 200010;
//
vector<pii> adj[MAX_N];
ll dist[MAX_N];
int N, M;
void Solve() {
    cin >> N >> M;
    for (int i = 2; i <= N; i++){
        adj[i].eb(i-1, 0);
        dist[i] = INT_MAX;
    }
    for (int i = 1, u, v, c; i <= M; i++){
        cin >> u >> v >> c;
        adj[u].eb(v, c);
    }
    
    queue<int> bfs;
    bfs.push(1);
    while (!bfs.empty()){
        int x = bfs.front(); bfs.pop();
        
        for (auto v:adj[x]){
            if (dist[v.F] > dist[x] + v.S){
                dist[v.F] = dist[x] + v.S;
                bfs.push(v.F);
            }
        }
    }
    cout << ((dist[N] != INT_MAX) ? dist[N]:-1);
}

signed main(){
    FAST_IO;
    int TC = 1; 
	// cin >> TC;
    while(TC--) Solve();
    return 0;
}
// 1 -> 0 0 0 0 1
// 2 -> 0 0 0 1 0
//