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;
vector<pair<int,int> > v[100005];
long long tree[400005];
void update(int node,int st,int en,int l,int r,long long val)
{
	if (en<l || st>r || r<l)
	return;
	if (l<=st && en<=r)
	{
		tree[node]=min(tree[node],val);
		return;
	}
	int mid=(st+en)/2;
	update(2*node,st,mid,l,r,val);
	update(2*node+1,mid+1,en,l,r,val);
}
long long query(int node,int st,int en,int idx)
{
	long long ans=tree[node];
	if (st!=en)
	{
		int mid=(st+en)/2;
		if (st<=idx && idx<=mid)
		ans=min(ans,query(2*node,st,mid,idx));
		else
		ans=min(ans,query(2*node+1,mid+1,en,idx));
	}
	return ans;
}
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	while (m--)
	{
		int l,r,c;
		scanf("%d%d%d",&l,&r,&c);
		v[l].push_back({r,c});
	}
	for (int i=1;i<=4*n;i++)
	tree[i]=1e18;
	update(1,1,n,1,1,0);
	for (int i=1;i<=n;i++)
	{
		long long cur=query(1,1,n,i);
		for (auto p:v[i])
		update(1,1,n,i,p.first,cur+p.second);
	}
	long long ans=query(1,1,n,n);
	if (ans==1e18)
	ans=-1;
	printf("%lld",ans);
}