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;

int main(){
    int n, m;
    cin >> n >> m;
    vector<pair<int,int64_t>> edges[100000];
    for(int i=0;i<m;++i){
        int l, r, c;
        cin >> l >> r >> c;
        edges[l-1].emplace_back(r-1, c);
    }
    for(int i=1;i<n;++i) edges[i].emplace_back(i-1, 0);

    priority_queue<pair<int64_t,int>> q;
    q.emplace(0, 0);
    bool checked[100000] = {0};
    while(!q.empty()){
        auto t = q.top();
        q.pop();
        if(checked[t.second]) continue;
        checked[t.second] = true;
        if(t.second == n-1){
            cout << (-t.first) << endl;
            return 0;
        }
        for(auto to : edges[t.second]){
            if(checked[to.first]) continue;
            q.emplace(t.first-to.second, to.first);
        }
    }
    cout << -1 << endl;

    return 0;
}

/*

*/