#include <bits/stdc++.h>
#define rep(i,n) for (int i=0; i<(n); ++i)
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;

const ll INF=1e18;
int n,m;
vector<P>edge[100010];

vector<ll> dijkstra(int x){
  vector<ll>dist(n+1,INF);
  priority_queue<P,vector<P>,greater<P>>q;
  dist[x]=0;
  q.push(P(0,x));
  while(!q.empty()){
    ll d=q.top().first;
    ll p=q.top().second;
    q.pop();
    if(dist[p]<d)continue;
    for(P next:edge[p]){
      ll to=next.first;
      ll co=next.second;
      if(dist[p]+co>=dist[to])continue;
      dist[to]=dist[p]+co;
      q.push(P(dist[to],to));
    }
  }
  return dist;
}

int main(){
  cin>>n>>m;
  rep(i,m){
    int l,r,c; cin>>l>>r>>c;
    edge[l].push_back(P(r,c));
  }
  rep(i,n-1){
    edge[i+1].push_back(P(i,0));
  }
  vector<ll>dist=dijkstra(1);
  if(dist[n]!=INF)cout<<dist[n]<<endl;
  else cout<<-1<<endl;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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>
#define rep(i,n) for (int i=0; i<(n); ++i)
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;

const ll INF=1e18;
int n,m;
vector<P>edge[100010];

vector<ll> dijkstra(int x){
  vector<ll>dist(n+1,INF);
  priority_queue<P,vector<P>,greater<P>>q;
  dist[x]=0;
  q.push(P(0,x));
  while(!q.empty()){
    ll d=q.top().first;
    ll p=q.top().second;
    q.pop();
    if(dist[p]<d)continue;
    for(P next:edge[p]){
      ll to=next.first;
      ll co=next.second;
      if(dist[p]+co>=dist[to])continue;
      dist[to]=dist[p]+co;
      q.push(P(dist[to],to));
    }
  }
  return dist;
}

int main(){
  cin>>n>>m;
  rep(i,m){
    int l,r,c; cin>>l>>r>>c;
    edge[l].push_back(P(r,c));
  }
  rep(i,n-1){
    edge[i+1].push_back(P(i,0));
  }
  vector<ll>dist=dijkstra(1);
  if(dist[n]!=INF)cout<<dist[n]<<endl;
  else cout<<-1<<endl;
}
```