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;
typedef pair<ll,ll> pp;
ll INF = 1e6;
ll n=INF,m;
vector<vector<pp>> graph(n,vector<pp>(0));


ll dijkstra(ll goal){
  priority_queue<pp,vector<pp>,greater<pp>> pq;
  pq.push(make_pair(0,0));
  vector<int> visited(n,0);
  while (!pq.empty()){
    auto now = pq.top();
    pq.pop();
    ll cost,node;
    tie(cost,node) = now;
    if(node == goal){
      return cost;
    }
    if(visited.at(node)==1){
      continue;
    }else{
      visited.at(node) = 1;
      for(auto next:graph.at(node)){
        ll next_cost,next_node;
        tie(next_cost,next_node) = next;
        pq.push(make_pair(cost+next_cost,next_node));
      }
    }
  }
  return -1;
}
      
  
int main() {
  cin >> n >> m;  
  for(int i=1;i<n;i++){
    graph.at(i).push_back(make_pair(0,i-1));
  }
  for(int i=0;i<m;i++){
    ll l,r,c;
    cin >> l >> r >> c;
    graph.at(l-1).push_back(make_pair(c,r-1));
  }
  cout << dijkstra(n-1) << endl;
}