Problem: 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.

Examples

Input

4 3
1 3 2
2 4 3
1 4 6


Output

5


Input

4 2
1 2 1
3 4 2


Output

-1


Input

10 7
1 5 18
3 4 8
1 3 5
4 7 10
5 9 8
6 10 5
8 10 3


Output

28

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

using namespace std ;

const int MAX = 1e5 + 10 ;

long long dist[MAX] ;

int main()
{
	ios_base::sync_with_stdio(0) ;
	cin.tie(0) ;
	int n , q;
	cin>>n>>q ;
	vector< vector< pair<long long , int> > >v(n+5) , v2(n+5);
	for(int i = 0 ; i < q ; ++i)
	{
		int x , y ;
		long long z ;
		cin>>x>>y>>z ;
		v[x].push_back({z , y}) ;
	}
	for(int i = 1 ; i <= n ; ++i)
		dist[i] = -1 ;
	dist[1] = 0 ;
	multiset< pair<long long , int> >s ;
	pair<long long , int>p ;
	for(int i = 1 ; i <= n ; ++i)
	{
		if(s.size() > 0)
		{
			p = *s.begin() ;
			dist[i] = p.first ;
		}
		if(dist[i] == -1)
			return cout<<-1<<"\n" , 0 ;
		for(auto &j : v2[i])
			s.erase(s.find(j)) ;
		for(auto &j : v[i])
		{
			s.insert({j.first+dist[i] , j.second}) ;
			v2[j.second].push_back({j.first+dist[i] , j.second}) ;
		}
	}
	return cout<<dist[n]<<"\n" , 0 ;
}