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:
// サンプルは合う
#include<iostream>
#include<vector>
#include<algorithm>
#include<set>
#include<queue>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;

int main(){
    int n, m;
    cin >> n >> m;
    vector<pii> vp[n+1];
    while(m-- > 0){
        int l, r, c;
        cin >> l >> r >> c;
        vp[l].push_back({r,c});
    }
    vector<ll> dp(n+1, 1ll<<60);
    dp[0] = 0;
    priority_queue<pair<ll,int>> pq;
    for(pii p : vp[1]){
        pq.push({-(dp[1-1]+p.second), p.first});
    }
    for(int i = 1; i <= n; i++){
        while(!pq.empty() && pq.top().second < i)  pq.pop();
        if(pq.empty()){
            cout << -1 << endl;
            return 0;
        }
        dp[i] = -pq.top().first;
        if(i != 1){
            for(pii p : vp[i]){
                pq.push({-(dp[i]+p.second), p.first});
            }
        }
    }
    cout << dp[n] << endl;
    return 0;
}