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

int n,m;
vector<P>to[100005];
vector<ll>dist(100005,LLONG_MAX);
ll solve(){
    priority_queue<P,vector<P>,greater<P>>pq;
    dist[0]=0;
    pq.push(P(0,0));
    while(pq.size()>0)
    {
        P cur=pq.top();pq.pop();
        ll d=cur.first;
        ll u=cur.second;
        if(d>dist[u])continue;

        for(P nx:to[u]){
            ll duv=nx.first;
            ll v=nx.second;
            if(dist[v]>d+duv){
                dist[v]=d+duv;
                pq.push(P(dist[v],v));
            }
        }
    }
     return dist[n-1]==LLONG_MAX?-1:dist[n-1];
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);

cin>>n>>m;
rep(i,m){
    ll l,r,c;
    cin>>l>>r>>c;
    --l,--r;
    to[l].push_back(make_pair(c,r));
}
for(int j=0;j<n;j++){
        to[j+1].push_back(make_pair(0,j));
    }
cout<<solve()<<endl;
return 0;
}
### Prompt

In Cpp, your task is to solve 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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
#define chmax(x,y) x = max(x,y)
#define chmin(x,y) x = min(x,y)

int n,m;
vector<P>to[100005];
vector<ll>dist(100005,LLONG_MAX);
ll solve(){
    priority_queue<P,vector<P>,greater<P>>pq;
    dist[0]=0;
    pq.push(P(0,0));
    while(pq.size()>0)
    {
        P cur=pq.top();pq.pop();
        ll d=cur.first;
        ll u=cur.second;
        if(d>dist[u])continue;

        for(P nx:to[u]){
            ll duv=nx.first;
            ll v=nx.second;
            if(dist[v]>d+duv){
                dist[v]=d+duv;
                pq.push(P(dist[v],v));
            }
        }
    }
     return dist[n-1]==LLONG_MAX?-1:dist[n-1];
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);

cin>>n>>m;
rep(i,m){
    ll l,r,c;
    cin>>l>>r>>c;
    --l,--r;
    to[l].push_back(make_pair(c,r));
}
for(int j=0;j<n;j++){
        to[j+1].push_back(make_pair(0,j));
    }
cout<<solve()<<endl;
return 0;
}
```