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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
const int N = 1e6+100;
struct node{
    int id;
    ll len;
};
ll dis[N];
vector<pair<int,ll> > G[N];
bool operator<(node a,node b){
    return a.len>b.len;
}
bool vis[N];
priority_queue<node> q;
int n,m;
void dij(int s){
    for(int i=1;i<=n;i++) dis[i]=(1e18);
    memset(vis,0,sizeof(vis));
    dis[s]=0;
    q.push({s,0});
    while(!q.empty()){
        node rt=q.top();q.pop();
        int u=rt.id;
        if(vis[u]) continue;
        vis[u]=1;
        for(auto V:G[u]){
            int v=V.first;
            ll w=V.second;
            if(dis[v]>dis[u]+w){
                dis[v]=dis[u]+w;
            }
            q.push({v,dis[v]});
        }
    }
}
int main(){
    cin>>n>>m;
    ll u,v,w;
    for(int i=2;i<=n;i++) G[i].push_back({i-1,0});
    for(int i=1;i<=m;i++){
        cin>>u>>v>>w;
        G[u].push_back({v,w});
    }
    dij(1);
    if(dis[n]==(1e18)){
        cout<<-1<<endl;
    }else{
        cout<<dis[n]<<endl;
    }
    return 0;
}