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<queue>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#define P pair<long long, long long>

using namespace std;
const long long int MOD=1e9+7;


int main(){
  long long n, m;
  vector<P> v[100009];
  priority_queue<P, vector<P>, greater<P>> Q1, Q2;
  cin >>n >> m;
  for(int i=0;i<m;i++){
    long long s, t, c;
    cin >> s >> t >> c;
    s--;t--;
    v[s].push_back(P(c, t));
  }
  for(int j=0;j<v[0].size();j++){
    Q1.push(v[0][j]);
    Q2.push(v[0][j]);
  }
  for(int i=1;i<n;i++){
    while(!Q1.empty() && Q1.top().second<i)Q1.pop();
    while(!Q2.empty() && Q2.top().second<i)Q2.pop();
    if(Q1.empty()){
      cout << -1 << endl;
      return 0;
    }
    long long tp = Q1.top().first;
    Q1.push(P(tp+Q2.top().first, Q2.top().second));
    for(int j=0;j<v[i].size();j++){
      Q1.push(P(tp + v[i][j].first, v[i][j].second));
      Q2.push(P(v[i][j]));
    }
  }
  cout << Q1.top().first << endl;
}