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;
const ll MAX = 1e17 + 7;
const ll N = 1e5+7;

struct aa{
    ll u,d;
    bool operator < (const aa & a) const{
        return d>a.d;
    }
};

vector<aa>arr[N];

bool vis[N];
ll n,m;
ll d[N];

void Dij(){
    fill(d+1,d+n+1,MAX);
    d[1]=0;
    priority_queue<aa>Q;
    Q.push({1,0});
    while(!Q.empty()){
        aa x=Q.top();
		Q.pop();
        ll u=x.u;
        if(vis[u])continue;
        vis[u]=1;
        for(ll i=0;i<arr[u].size();i++){
            aa e=arr[u][i];
            if(d[e.u]>d[u]+e.d){
                d[e.u]=d[u]+e.d;
                Q.push({e.u,d[e.u]});
            }
        }
    }
}

int main(){
	ll a,b,w;		
	scanf("%lld%lld",&n,&m);
	for(ll i=2;i<=n;i++){
		arr[i].push_back({i-1,0});
	}
	for(ll i=0;i<m;i++){
		scanf("%lld%lld%lld",&a,&b,&w);
		arr[b].push_back({a,w});
		arr[a].push_back({b,w});
	}
	Dij();
	if(d[n]==MAX) d[n]=-1;
	cout<<d[n]<<endl;
	
	
	return 0;
}