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>
#define MAX (ll)(2e5+5)
#define MOD (ll)(998244353)
#define INF (ll)(1e18 + 3)
#define PI (double)(3.14159265)

using namespace std;
using ll = long long;

int main () {
	int n;
	cin >> n;
	pair<int, int> aa[n], ar[n], br[n];
	int ga[n], gb[n];
	for (int i = 0; i < n; i++) {
		cin >> ar[i].first;
		ar[i].second = i;
		aa[i].first = ar[i].first;
	}	
	for (int i = 0; i < n; i++) {
		cin >> br[i].first;
		br[i].second = i;
		aa[i].second = br[i].first;
	}	
	sort (ar, ar + n);
	sort (br, br + n);
	for (int i = 0; i < n; i++) {
		if (ar[i].first > br[i].first) {
			cout << "No\n";
			return 0;
		}
	}
	for (int i = 0; i < n; i++) {
		if (i && br[i - 1].first >= ar[i].first) {		
			cout << "Yes\n";
			return 0;
		}
		ga[ar[i].second] = i;
		gb[br[i].second] = i;
	}
	set <pair<int, int>> sp;
	for (int i = 0; i < n; i++) {
		if (ga[i] == gb[i] || (sp.count (make_pair (gb[i], ga[i])))) {
			cout << "Yes\n";
			return 0;
		}
		sp.insert (make_pair (ga[i], gb[i]));
	}
	cout << "No\n";
}