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;
typedef long long ll;
typedef pair<ll,int>P;
const int N=1e5+5,M=131072;
int n,m,l[N],r[N],c[N],i;
bool b[N];
vector<int>t[N*3];
void ins(int l,int r,int x){
	for(l+=M-1,r+=M+1;l^r^1;l>>=1,r>>=1){
		if(!(l&1))t[l^1].push_back(x);
		if(r&1)t[r^1].push_back(x);
	}
}
set<int>s;
priority_queue<P,vector<P>,greater<P>>q;
inline void work(int x,ll d){
	vector<int>ve;
	for(x+=M;x;x>>=1)ve.insert(ve.end(),t[x].begin(),t[x].end()),t[x].clear();
	for(int x:ve)if(!b[x])b[x]=1,q.push({d+c[x],x});
}
int main(){
	scanf("%d%d",&n,&m);
	for(i=1;i<=m;++i)scanf("%d%d%d",l+i,r+i,c+i),ins(l[i],r[i],i);
	for(i=1;i<=n;++i)s.insert(i);
	work(1,0);
	for(;!q.empty();){
		ll d;int x;tie(d,x)=q.top();q.pop();
		if(r[x]==n){printf("%lld\n",d);return 0;}
		for(auto it=s.lower_bound(l[x]);it!=s.end() && *it<=r[x];s.erase(it++))work(*it,d);
	}
	puts("-1");
	return 0;
}