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 JAVA):

Solution:
import java.util.*;

public class Main{

    static Scanner sc = new Scanner(System.in);
    static int n = sc.nextInt();
    static int m = sc.nextInt();
    static List<Node>[] g = new ArrayList[n];
    static long[] d = new long[n];

    public static void main(String[] args){
        d[0] = 0;
        for (int i=0; i<n; i++){
            g[i] = new ArrayList<>();
        }

        for (int i=0; i<n; i++) {
            g[i] = new ArrayList<Node>();
            if (i!=0) g[i].add(new Node(i-1, 0));

        }
        for (int i=0; i<m; i++){
            int l = sc.nextInt()-1;
            int r = sc.nextInt()-1;
            int c = sc.nextInt();
            g[l].add(new Node(r, c));
        }

        System.out.println(dijkstra());

    }

    public static long dijkstra(){
        Arrays.fill(d, Long.MAX_VALUE);
        d[0] = 0;
        PriorityQueue<Node> pq = new PriorityQueue<>();
        pq.add(new Node(0, 0));

        while (!pq.isEmpty()){
            Node nd = pq.poll();
            if (nd.node==n-1) return nd.cost;
            if (nd.cost > d[nd.node]) continue;

            for (Node next: g[nd.node]){
                if (d[next.node] > nd.cost+next.cost){
                    d[next.node] = nd.cost+next.cost;
                    pq.add(new Node(next.node, nd.cost+next.cost));
                }
            }

        }
        return -1;
    }

    public static class Node implements Comparable<Node>{
        int node;
        long cost;
        public Node(int n, long c){
            this.node = n;
            this.cost = c;
        }
        public int compareTo(Node nd){
            if (this.cost<nd.cost) return -1;
            if (this.cost>nd.cost) return 1;
            return 0;
        }
    }

}