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 finish(x) return cout << x << endl, 0
#define ll long long

const int N = 100001;

int n, m;
vector <pair <int, int> > v[N];

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> m;
    for(int i = 0 ; i < m ; i++){
        int l, r, c;
        cin >> l >> r >> c;
        v[l].push_back(make_pair(i, c));
        v[r].push_back(make_pair(i, -1));
    }
    multiset <ll> s;
    set <pair <int, ll> > g;
    ll dp = 0;
    for(int i = 1 ; i <= n ; i++){
        if(i > 1){
            if(s.size() == 0) finish(-1);
            else dp = *s.begin();
        }
        for(auto &j : v[i]){
            if(j.second != -1){
                s.insert(dp + j.second);
                g.insert(make_pair(j.first, dp + j.second));
            }
            else{
                auto it = g.lower_bound(j);
                s.erase(s.find((*it).second));
                g.erase(it);
            }
        }
    }
    cout << dp << endl;
}