#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<ll,ll> P;
const ll INF = 1LL << 50;

vector<vector<P>> to; vector<ll> to_dist;

void Dijkstra(ll v){
  
  to_dist[v] = 0; 
  
  priority_queue<P, vector<P>, greater<P>> Q; 
  Q.push({0,v}); 
  
  while(!Q.empty()){
   
    P p = Q.top(); Q.pop();
    ll s = p.first; ll t = p.second; 
      for(auto ne: to[t]){
        ll x = ne.first; ll y = ne.second;
        
        if( to_dist[x] > to_dist[t] + y){
          to_dist[x] = to_dist[t] + y; Q.push({to_dist[x], x});}
      }
  }

}


int main(){
  ll N,M; cin >> N >> M;
  to.resize(N); to_dist = vector<ll> (N,INF);
  
  for(ll i = 0; i < N-1; i++){  to[i+1].push_back({i,0});}
  
  for(ll i = 0; i < M; i++){
    ll L,R,C; cin >> L >> R >> C; L--; R--; 
    to[L].push_back({R,C}); to[R].push_back({L,C});
  }
  
  Dijkstra(0);
  ll ans = to_dist[N-1]; 
  if( ans == INF){ ans = -1;} 
  
  cout << ans << endl; return 0;
}
    
  
 
  
### Prompt

Please provide a Cpp coded solution to the problem described below:

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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<ll,ll> P;
const ll INF = 1LL << 50;

vector<vector<P>> to; vector<ll> to_dist;

void Dijkstra(ll v){
  
  to_dist[v] = 0; 
  
  priority_queue<P, vector<P>, greater<P>> Q; 
  Q.push({0,v}); 
  
  while(!Q.empty()){
   
    P p = Q.top(); Q.pop();
    ll s = p.first; ll t = p.second; 
      for(auto ne: to[t]){
        ll x = ne.first; ll y = ne.second;
        
        if( to_dist[x] > to_dist[t] + y){
          to_dist[x] = to_dist[t] + y; Q.push({to_dist[x], x});}
      }
  }

}


int main(){
  ll N,M; cin >> N >> M;
  to.resize(N); to_dist = vector<ll> (N,INF);
  
  for(ll i = 0; i < N-1; i++){  to[i+1].push_back({i,0});}
  
  for(ll i = 0; i < M; i++){
    ll L,R,C; cin >> L >> R >> C; L--; R--; 
    to[L].push_back({R,C}); to[R].push_back({L,C});
  }
  
  Dijkstra(0);
  ll ans = to_dist[N-1]; 
  if( ans == INF){ ans = -1;} 
  
  cout << ans << endl; return 0;
}
    
  
 
  
```