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<algorithm>
#include<vector>
#include<queue>
#include<set>
#include<unordered_map>
using namespace std;
typedef long long ll;
#define chmax(a,b) a=max(a,b)
#define chmin(a,b) a=min(a,b)
#define mod 1000000007
#define mad(a,b) a=(a+b)%mod
#define N 1000000
vector<pair<ll,ll> >g[N];
typedef pair<ll,ll> P;
priority_queue<P,vector<P>,greater<P> >Q;
ll d[N];
int main(){
    cin.tie(0);
    ios::sync_with_stdio(0);
    ll n,m;
    cin>>n>>m;
    for(int i=1;i<n;i++){
	g[i+1].push_back(make_pair(i,0));
    }
    for(int i=0;i<m;i++){
	ll a,b,c;
	cin>>a>>b>>c;
	g[a].push_back(make_pair(b,c));
    }
    for(int i=1;i<=n;i++)d[i]=1e17;
    Q.push(make_pair(0,1));
    while(!Q.empty()){
	ll x=Q.top().second;
	ll cost=Q.top().first;
	Q.pop();
	if(d[x]<=cost)continue;
	d[x]=cost;
	for(auto y:g[x]){
	    Q.push(make_pair(cost+y.second,y.first));
	}
    }
    if(d[n]==1e17)d[n]=-1;
    cout<<d[n]<<endl;
}