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,int> P;

vector<vector<P>> to;

const int MAX_N = 101000;
const ll INF = 1LL << 50;

ll dist[MAX_N];

void Dijkstra(int v){
  //vが始点
  dist[v] = 0;
  priority_queue<P,vector<P>,greater<P>> Q;
  // {距離、ノード}の組みを保持
  Q.push({0,v});

  while(!Q.empty()){
    P p = Q.top();  Q.pop();
    ll load = p.first; int pos = p.second;
    
    for(auto next : to[pos]){ 
      int u = next.first; ll cost = next.second;
      if(dist[u] <= load+cost){ continue;} 
      else{ dist[u] = load+cost;
           
           Q.push({dist[u],u});
          }
    } // for-とじ
  }// whileとじ
  
  return ;
}
    

int main(){ 
  int N,M; cin >> N >> M;
  to.resize(N);
  
  for(int i = 0; i < N; i++){ dist[i] = INF;}
  
  for(int i = 0; i < M; i++){
    int L,R; ll C; cin >> L >> R >> C; L--; R--;
    to[L].push_back({R,C});
  }
  
  for(int i = 0; i < N-1; i++){
    to[i+1].push_back({i,0});
  }
  
  //隣接リストto上で、0からN-1まで動くのにかかる時間を見つける
  
  Dijkstra(0);
  
  ll res = dist[N-1];
  if( res == INF){ res = -1;}
  
  cout << res << endl;
  
  return 0;
}