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;

#define int long long

typedef pair<int, int> P;

int dist[100010];

vector<P> vec[100010];

signed main(){
    int n, m;
    cin >> n >> m;

    priority_queue<P, vector<P>, greater<P>> que;

    for(int i = 0;i < m;i++){
        int l, r, c;
        cin >> l >> r >> c;
        l--;
        r--;

        vec[l].push_back({r, c});
    }


    memset(dist, -1, sizeof(dist));

    dist[0] = 0;
    for(auto e : vec[0]){
        que.push({e.second, e.first});
    }


    int now = 0;
    while(!que.empty()){
        int cost = que.top().first;
        int r = que.top().second;
        que.pop();
        if(now < r){
            for(int i = now+1;i <= r;i++){
                dist[i] = cost;
                for(auto e : vec[i]){
                    que.push({e.second+cost, e.first});
                }
            }
            now = r;
        }
    }

    cout << dist[n-1] << endl;

    return 0;
}