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;
#define rep(i,a) for(int i=0;i<(a);i++)
const ll MOD=1000000007;
//const ll MOD=998244353;

int N,M;
ll L[101010],R[101010],C[101010];
vector<pair<ll,int>> V;
set<pair<ll,int>> st;
ll d[101010];
const ll INF=1e15;

int main(){
  cin>>N>>M;
  V.resize(M);
  rep(i,M){
    cin>>L[i]>>R[i]>>C[i]; L[i]--,R[i]--;
    V[i]={L[i],i};
  }
  sort(V.begin(),V.end());
  V.push_back({1e9,0});
  rep(i,N) d[i]=INF;
  d[0]=0;
  int ind=0;
  rep(i,N){
    while(st.size()){
      auto it=st.begin();
      if(it->second<=i){
        st.erase(it);
      }else{
        d[i]=min(d[i],it->first);
        break;
      }
    }
    if(d[i]==INF){
      cout<<-1<<endl;
      return 0;
    }
    while(V[ind].first==i){
      int j=V[ind++].second;
      if(d[R[j]]>d[i]+C[j]){
        d[R[j]]=d[i]+C[j];
        st.insert({d[R[j]],R[j]});
      }
    }
  }
  cout<<d[N-1]<<endl;
  return 0;
}