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;

int main(){
  int N,M;
  cin>>N>>M;
  vector<tuple<int,int,int64_t>> p(M+1);
  for(int i=0;i<M;i++){
    int a,b;
    int64_t c;
    cin>>a>>b>>c;
    a--,b--;
    p[i]=make_tuple(a,b,c);
  }
  p[M]=make_tuple(167167,167167,167167);
  sort(p.begin(),p.end());
  vector<int64_t> l(N);
  priority_queue<int64_t,vector<int64_t>,greater<int64_t>> pq;
  priority_queue<int64_t,vector<int64_t>,greater<int64_t>> pqo;
  pq.push(0);
  int a=0;
  vector<vector<int64_t>> G(N);
  G[0].push_back(0);
  for(int i=0;i<N;i++){
    l[i]=pq.top();
    if(i==N-1){
      break;
    }
    while(get<0>(p[a])==i){
      int64_t D=get<2>(p[a])+l[i];
      G[get<1>(p[a])].push_back(D);
      pq.push(D);
      a++;
    }
    for(auto x:G[i]){
      pqo.push(x);
    }
    while(pq.top()==pqo.top()&&!pq.empty()&&!pqo.empty()){
      pq.pop();
      pqo.pop();
    }
    if(pq.empty()){
      l[N-1]=-1;
      break;
    }
  }
  cout<<l[N-1]<<endl;
    
}