#include<bits/stdc++.h>
using namespace std;
using ll=long long;
static const ll INF=1000000000000000;
ll dat[400005];
ll n;
void init(ll n_){
  n=1;
while(n<n_)n*=2;
  for(ll i=0;i<2*n-1;i++)
    dat[i]=INF;
}
void update(ll k,ll a){
k+=n-1;
  dat[k]=a;
  while(0<k){
  k=(k-1)/2;
    dat[k]=min(dat[2*k+1],dat[2*k+2]);
  }
      }
ll query(ll a,ll b,ll k,ll l,ll r){
if(b<=l || r<=a)
  return INF;
  else if(a<=l && r<=b)
    return dat[k];
  else{
  ll val=query(a,b, 2*k+1,l,(l+r)/2);
    ll var=query(a,b,2*k+2,(l+r)/2,r);
    return min(val,var);
  }
}
ll N,M;
int main(){
 cin>>N>>M;
  init(N);
  vector<pair<pair<ll,ll>,ll>>A(M);
  for(int i=0;i<M;i++){
    ll l,r,c;cin>>l>>r>>c;l--;r--;
    pair<ll,ll>p(r,l);
    pair<pair<ll,ll>,ll>q(p,c);
    A[i]=q;
     }sort(A.begin(),A.end());
  update(0,0);
  for(int i=0;i<M;i++){
  ll r=A[i].first.first;
    ll l=A[i].first.second;
    ll c=A[i].second;
  ll m=query(l,r,0,0,n);
    if(m==INF)
      continue;
    if(m+c<dat[r+n-1])
      update(r,m+c);
  }if(dat[N-1+n-1]<INF)
  cout<<dat[N-1+n-1]<<endl;
  else
    cout<<-1<<endl;
  
 
     return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following 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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
static const ll INF=1000000000000000;
ll dat[400005];
ll n;
void init(ll n_){
  n=1;
while(n<n_)n*=2;
  for(ll i=0;i<2*n-1;i++)
    dat[i]=INF;
}
void update(ll k,ll a){
k+=n-1;
  dat[k]=a;
  while(0<k){
  k=(k-1)/2;
    dat[k]=min(dat[2*k+1],dat[2*k+2]);
  }
      }
ll query(ll a,ll b,ll k,ll l,ll r){
if(b<=l || r<=a)
  return INF;
  else if(a<=l && r<=b)
    return dat[k];
  else{
  ll val=query(a,b, 2*k+1,l,(l+r)/2);
    ll var=query(a,b,2*k+2,(l+r)/2,r);
    return min(val,var);
  }
}
ll N,M;
int main(){
 cin>>N>>M;
  init(N);
  vector<pair<pair<ll,ll>,ll>>A(M);
  for(int i=0;i<M;i++){
    ll l,r,c;cin>>l>>r>>c;l--;r--;
    pair<ll,ll>p(r,l);
    pair<pair<ll,ll>,ll>q(p,c);
    A[i]=q;
     }sort(A.begin(),A.end());
  update(0,0);
  for(int i=0;i<M;i++){
  ll r=A[i].first.first;
    ll l=A[i].first.second;
    ll c=A[i].second;
  ll m=query(l,r,0,0,n);
    if(m==INF)
      continue;
    if(m+c<dat[r+n-1])
      update(r,m+c);
  }if(dat[N-1+n-1]<INF)
  cout<<dat[N-1+n-1]<<endl;
  else
    cout<<-1<<endl;
  
 
     return 0;
}

```