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;
using ll=long long;
#define rep(i,n) for(ll i=0;i<n;i++)
const ll INF=1e15;
const ll MOD=1000000007;

int main(){
    ll N,M;cin>>N>>M;
    vector<pair<ll,ll>> info[N];
    rep(i,M){
        ll L,R,C;cin>>L>>R>>C;
        L--;R--;
        info[L].push_back(make_pair(R,C));
    }
    rep(i,N-1)info[i+1].push_back(make_pair(i,0));
    ll sum[N];
    rep(i,N){
        if(!i)sum[i]=0;
        else sum[i]=INF;
    }
    priority_queue<pair<ll,ll>,vector<pair<ll,ll>>,greater<pair<ll,ll>>> que;
    que.push(make_pair(0,0));
    while(que.size()){
        ll d=que.top().first,v=que.top().second;
        que.pop();
        for(auto i:info[v]){
            ll nv=i.first;
            ll nc=i.second;
            if(d+nc<sum[nv]){
                sum[nv]=d+nc;
                que.push(make_pair(sum[nv],nv));
            }
        }
    }
    if(sum[N-1]==INF)cout<<-1<<endl;
    else cout<<sum[N-1]<<endl;
    return 0;
}