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 MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,ll> P;

class segtree{
public:
	static const int N=1<<17;
	ll dp[1<<18];
	segtree(){
		for(int i=0;i<N*2;i++){
			dp[i]=(1LL<<60);
		}
	}
	void update(int k,ll v){
		k+=N-1;
		dp[k]=min(v,dp[k]);
		while(k>0){
			k=(k-1)/2;
			dp[k]=min(dp[k*2+1],dp[k*2+2]);
		}
	}

	ll query(int a,int b,int k=0,int l=0,int r=N){
		if(b<=l || r<=a)return (1LL<<60);
		if(a<=l && r<=b)return dp[k];
		int mid=(l+r)/2;
		ll vl=query(a,b,k*2+1,l,mid);
		ll vr=query(a,b,k*2+2,mid,r);
		return min(vl,vr);
	}
};

int n,m;
vector<P> G[100005]; 
segtree seg;

int main(void){
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++){
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		a--;
		b--;
		G[a].push_back(P(b,c));
	}
	seg.update(0,0);
	for(int i=0;i<n-1;i++){
		ll ic=seg.query(i,n);
		for(int j=0;j<G[i].size();j++){
			P p=G[i][j];
			int to=p.first;
			ll c=p.second;
			seg.update(to,ic+c);
		}
	}
	ll ans=seg.query(n-1,n);
	printf("%lld\n",ans==(1LL<<60)?-1:ans);
	return 0;
}