#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int nmax=100005;
struct edge{ll to;ll cost;};
vector<edge> tree[nmax];
ll d[nmax];
ll n;
void dijkstra(ll s){
  priority_queue<pair<ll,ll>,vector<pair<ll,ll>>,greater<pair<ll,ll>>> pq;
  fill(d,d+n,1e18);
  d[s]=0;
  pq.push(make_pair(0,s));
  while(!pq.empty()){
    pair<ll,ll> p=pq.top();
    pq.pop();
    ll v=p.second;
    if(d[v]<p.first){
      continue;
    }
    for(auto x:tree[v]){
      edge e=x;
      if(d[e.to]>d[v]+e.cost){
        d[e.to]=d[v]+e.cost;
        pq.push(make_pair(d[e.to],e.to));
      }
    }
  }
}
int main(){
  ll m;cin >> n >> m;
  for(int i=1;i<n;i++){
    tree[i].push_back(edge{i-1,0});
  }
  for(int i=0;i<m;i++){
    ll l,r,c;cin >> l >> r >> c;
    l--,r--;
    tree[l].push_back(edge{r,c});
  }
  dijkstra(0);
  if(d[n-1]==1e18){
    cout << -1 << endl;
  }
  else{
    cout << d[n-1] << endl;
  }
}
### Prompt

Develop a solution in CPP 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;
const int nmax=100005;
struct edge{ll to;ll cost;};
vector<edge> tree[nmax];
ll d[nmax];
ll n;
void dijkstra(ll s){
  priority_queue<pair<ll,ll>,vector<pair<ll,ll>>,greater<pair<ll,ll>>> pq;
  fill(d,d+n,1e18);
  d[s]=0;
  pq.push(make_pair(0,s));
  while(!pq.empty()){
    pair<ll,ll> p=pq.top();
    pq.pop();
    ll v=p.second;
    if(d[v]<p.first){
      continue;
    }
    for(auto x:tree[v]){
      edge e=x;
      if(d[e.to]>d[v]+e.cost){
        d[e.to]=d[v]+e.cost;
        pq.push(make_pair(d[e.to],e.to));
      }
    }
  }
}
int main(){
  ll m;cin >> n >> m;
  for(int i=1;i<n;i++){
    tree[i].push_back(edge{i-1,0});
  }
  for(int i=0;i<m;i++){
    ll l,r,c;cin >> l >> r >> c;
    l--,r--;
    tree[l].push_back(edge{r,c});
  }
  dijkstra(0);
  if(d[n-1]==1e18){
    cout << -1 << endl;
  }
  else{
    cout << d[n-1] << endl;
  }
}
```