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>
#define pb push_back
#define mp make_pair
#define int long long
#define inf 1000000007
#define LINF 100000000000000007LL
using namespace std;
typedef pair<int,int> P;
struct edge{int to, cost; };
int n;
vector< edge > g[114514];
int d[114514];
void dijkstra( int s ){
	priority_queue< P, vector<P>, greater<P> > que;
	for(int i=1;i<=n;i++){
		d[i] = LINF;
	}
	d[s] = 0;
	que.push( P(0,s) );
	while( !que.empty() ){
		P p = que.top(); que.pop();
		int v = p.second;
		if( d[v] < p.first ) continue;
		for(int i=0;i<g[v].size();i++){
			edge e = g[v][i];
			if( d[e.to] > d[v] + e.cost ){
				d[e.to] = d[v] + e.cost;
				que.push(P(d[e.to],e.to));
			}
		}
	}
}
signed main(){
	cin>>n;
	int m;
	cin>>m;
	for(int i=0;i<m;i++){
		int l,r,c;
		cin>>l>>r>>c;
		g[l].pb( (edge){r,c} );
	}
	for(int i=2;i<=n;i++){
		g[i].pb( (edge){i-1,0} );
	}
	dijkstra(1);
	if( d[n] == LINF )d[n]=-1;
	cout<<d[n]<<endl;
	return 0;
}