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>
using namespace std;
int N_MAX = 100001;
long long INF=123456789123456789;
typedef pair<int,long long> P;
typedef pair<P,int> PP;
int main()
{
  int n,m,l,r,i;
  long long c;
  long long res[N_MAX];
  PP pp;
  priority_queue<PP,vector<PP>,greater<PP>> q;
  cin >> n >> m;
  res[1]=0;
  fill(res+2,res+n+1,INF);
  for(i=0;i<m;i++){
    cin >> l >> r >> c;
    q.push(PP(P(l,c),r));
  }
  pp=q.top();
  if(pp.first.first>1){
    cout << "-1" << endl;
    return 0;
  }
  while(!q.empty()){
    pp=q.top(); q.pop();
    l=pp.first.first;
    c=pp.first.second;
    r=pp.second;
    for(i=r;i>l;i--)
      if(res[l]+c<res[i]){
        res[i]=res[l]+c;
      }else{
        break;
      }
  }
  if(res[n]==INF)
    cout << "-1" << endl;
  else
    cout << res[n] << endl;
  return 0;
}