#include <bits/stdc++.h>
#define rep(i,l,r) for (int i=l,i##end=r;i<=i##end;++i)
#define per(i,r,l) for (int i=r,i##end=l;i>=i##end;--i)
#define ll long long
#define pr pair<int,int>
#define mp make_pair
#define fi first
#define se second
using namespace std;
const int N=1e6+6; const ll INF=1e18;
vector< pr > G[N];
int st[N],tp;
ll f[N];
int main(){
    int n,m; cin>>n>>m;
    rep(i,1,m){
        int l,r,c; cin>>l>>r>>c;
        G[r].push_back(mp(l,c));
    }
    f[0]=INF,f[1]=0; st[tp=1]=1;
    rep(i,2,n){
        f[i]=INF;
        rep(j,0,G[i].size()-1){
            int l=G[i][j].fi,v=G[i][j].se;
            f[i]=min(f[i],v+f[*lower_bound(st+1,st+1+tp,l)]);
        }
        while (tp&&f[i]<f[st[tp]]) --tp;
        st[++tp]=i;
    }
    cout<<(f[n]==INF?-1:f[n])<<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>
#define rep(i,l,r) for (int i=l,i##end=r;i<=i##end;++i)
#define per(i,r,l) for (int i=r,i##end=l;i>=i##end;--i)
#define ll long long
#define pr pair<int,int>
#define mp make_pair
#define fi first
#define se second
using namespace std;
const int N=1e6+6; const ll INF=1e18;
vector< pr > G[N];
int st[N],tp;
ll f[N];
int main(){
    int n,m; cin>>n>>m;
    rep(i,1,m){
        int l,r,c; cin>>l>>r>>c;
        G[r].push_back(mp(l,c));
    }
    f[0]=INF,f[1]=0; st[tp=1]=1;
    rep(i,2,n){
        f[i]=INF;
        rep(j,0,G[i].size()-1){
            int l=G[i][j].fi,v=G[i][j].se;
            f[i]=min(f[i],v+f[*lower_bound(st+1,st+1+tp,l)]);
        }
        while (tp&&f[i]<f[st[tp]]) --tp;
        st[++tp]=i;
    }
    cout<<(f[n]==INF?-1:f[n])<<endl;
    return 0;
}

```