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<iostream>
#include<cstdio>
#include<queue>
#define ll long long
using namespace std;
int n,m,x,y,head[1000010],cnt=0;
ll z,dis[1000010];
struct Node{
	int u;
	ll dis;
	bool operator <(const Node &rhs)const{
		return dis>rhs.dis;
	}
};
priority_queue<Node>q;
struct Edge{
	int v,nxt;
	ll w;
}e[1000010];
void addEdge(int u,int v,ll w){
	e[++cnt].v=v;
	e[cnt].w=w;
	e[cnt].nxt=head[u];
	head[u]=cnt;
}
int main(){
	scanf("%d%d",&n,&m);
	for(register int i=2;i<=n;i++)addEdge(i,i-1,0);
	for(register int i=1;i<=m;i++){
		scanf("%d%d%d",&x,&y,&z);
		addEdge(x,y,z);
	}
	for(register int i=1;i<=n;i++)dis[i]=1e18;
	dis[1]=0;
	q.push((Node){1,0});
	while(!q.empty()){
		Node now=q.top();q.pop();
		if(now.dis!=dis[now.u])continue;
		for(register int i=head[now.u];i;i=e[i].nxt){
			if(dis[e[i].v]>dis[now.u]+e[i].w){
				dis[e[i].v]=dis[now.u]+e[i].w;
				q.push((Node){e[i].v,dis[e[i].v]});
			}
		}
	}
	printf("%lld",dis[n]<1e18?dis[n]:-1);
	return 0;
}