#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;
#define ll long long
#define rep(i,n) for(int i=0;i<(n);i++)
#define repp(i,n) for(int i=1;i<=n;i++)

//負の辺がない場合
typedef pair<ll,int> P;
const int MAX = 100100;
const ll INF = 1e15;
struct edge{
    int to,cost;
};
int v;
vector<vector<edge>> G(MAX);
ll d[MAX];

void dijkstra(int s){
    priority_queue<P,vector<P>,greater<P>> pq;
    rep(i,v) d[i] = INF;
    d[s]=0;
    pq.push({0,s});
    while(!pq.empty()){
        P p = pq.top(); pq.pop();
        int v = p.second;
        if(d[v]<p.first) continue;
        for(edge u:G[v]){
            if(d[u.to]>d[v]+u.cost){
                d[u.to]=(ll)d[v]+u.cost;
                pq.push({d[u.to],u.to});
            }
        }
    }
}

int main(){
    cin >> v;
    int n; cin >> n;
    rep(i,n){
        int l,r,c; cin >> l >> r >> c;
        l--; r--;
        G[l].push_back({r,c});
    }
    rep(i,v) if(i+1<n) G[i+1].push_back({i,0});
    dijkstra(0);
    if(d[v-1]<INF) cout << d[v-1];
    else cout << -1;
}

### 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 <iostream>
#include <vector>
#include <algorithm>
#include <queue>
using namespace std;
#define ll long long
#define rep(i,n) for(int i=0;i<(n);i++)
#define repp(i,n) for(int i=1;i<=n;i++)

//負の辺がない場合
typedef pair<ll,int> P;
const int MAX = 100100;
const ll INF = 1e15;
struct edge{
    int to,cost;
};
int v;
vector<vector<edge>> G(MAX);
ll d[MAX];

void dijkstra(int s){
    priority_queue<P,vector<P>,greater<P>> pq;
    rep(i,v) d[i] = INF;
    d[s]=0;
    pq.push({0,s});
    while(!pq.empty()){
        P p = pq.top(); pq.pop();
        int v = p.second;
        if(d[v]<p.first) continue;
        for(edge u:G[v]){
            if(d[u.to]>d[v]+u.cost){
                d[u.to]=(ll)d[v]+u.cost;
                pq.push({d[u.to],u.to});
            }
        }
    }
}

int main(){
    cin >> v;
    int n; cin >> n;
    rep(i,n){
        int l,r,c; cin >> l >> r >> c;
        l--; r--;
        G[l].push_back({r,c});
    }
    rep(i,v) if(i+1<n) G[i+1].push_back({i,0});
    dijkstra(0);
    if(d[v-1]<INF) cout << d[v-1];
    else cout << -1;
}

```