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>
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
struct Edge{int to,cost;};
struct Node{
    int v,d;
    bool operator<(const Node&right)const{return d<right.d;}
    bool operator>(const Node&right)const{return d>right.d;}
};
int n,m,l,r,c,d[100009];
vector<Edge> e[100009];

signed main(){
    cin>>n>>m;
    for(int i=1;i<n;i++){
        e[i+1].push_back({i,0});
        d[i+1]=1e18;
    }
    rep(i,m){
        cin>>l>>r>>c;
        e[l].push_back({r,c});
    }
    priority_queue<Node,vector<Node>,greater<Node>> que;
    que.push({1,0});
    while(!que.empty()){
        Node V=que.top();
        que.pop();
        if(d[V.v]<V.d)continue;
        rep(i,e[V.v].size()){
            if(d[e[V.v][i].to]>V.d+e[V.v][i].cost){
                d[e[V.v][i].to]=V.d+e[V.v][i].cost;
                que.push({e[V.v][i].to,d[e[V.v][i].to]});
            }
        }
    }
    cout<<(d[n]<1e18?d[n]:-1)<<endl;
}