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; 
#define MOD 1000000007
#define f(i, n) for(ll i=0; i < (n); i++)
typedef long long int ll;
typedef vector<ll> llvec;
typedef vector<double> dvec;
typedef pair<int, int> P;
const ll mk = 1e5 ; 
int main(){
  int N , M ; cin >> N >> M ;
  vector<vector<P> > v(N);
  vector<ll> cost(N, 1e18);
  f(i,N-2){
    v[i+2].push_back(make_pair(i+1, 0));
  }
  f(i, M){
    ll l, r, c;
    cin >> l >> r >> c;
    l--;r--;
    v[l].push_back(make_pair(r,c));
  }
  
  cost[0]=0;
  priority_queue<P, vector<P>, greater<P> > Queue;
  llvec check(N, 0);
  Queue.push(make_pair(0, 0));
  while(!Queue.empty()){
    P p =Queue.top();
    Queue.pop();
    ll from = p.second;
    
    if(check[from]!=0)continue;
    check[from]=1;
    ll vnum = v[from].size();
    f(i, vnum){
      ll to = v[from][i].first;
      ll c = v[from][i].second;
      if(cost[to]>cost[from]+c){
        cost[to]=cost[from]+c;
        Queue.push(make_pair(cost[to], to));
      }
    }
  }
  
  cout << (cost[N-1]==1e18?-1:cost[N-1]);
  
  return 0;
}