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;

using i64 = int64_t;

constexpr i64 INF = 1e18;

struct Edge{
    i64 from, to, cost;
};

struct State{
    i64 num, cost;
    bool operator<(const State &obj)const{
        return cost > obj.cost;
    }
};

using Graph = vector<vector<Edge>>;

int main(){
    int n, m;
    cin >> n >> m;
    Graph graph(n+1, vector<Edge>());
    for(int i=0;i<m;++i){
        i64 l, r, c;
        cin >> l >> r >> c;
        graph[l].push_back({l, r, c});
    }
    for(int i=n;i>0;--i){
        graph[i].push_back({i, i-1, 0});
    }

    vector<i64> mini(n+1, INF);
    mini[1] = 0;
    priority_queue<State> que;
    que.push({1, 0});
    while(!que.empty()){
        auto cur = que.top(); que.pop();
        if(cur.num == n){
            cout << cur.cost << endl;
            return 0;
        }
        for(auto e: graph[cur.num]){
            i64 new_cost = cur.cost + e.cost;
            if(new_cost >= mini[e.to])continue;
            mini[e.to] = new_cost;
            que.push({e.to, new_cost});
        }
    }
    cout << -1 << endl;

    return 0;
}