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;
typedef long long ll;
typedef pair<ll,ll> P;
typedef pair<P,ll> Q;

ll N,M;
vector<vector<P>> G;
ll d[200000];

int main(){
    cin >> N >> M;
    G.resize(N);
    for(int i=1;i<N;i++) G[i].push_back({i-1,0});
    for(int i=0;i<M;i++){
        ll l,r,c;
        cin >> l >> r >> c;
        G[l-1].push_back({r-1,c});
    }
    priority_queue<P,vector<P>,greater<P>> pq;
    pq.push({0,0});
    for(int i=0;i<N;i++) d[i] = -1;
    while(d[N-1] < 0 && !pq.empty()){
        P p = pq.top();
        pq.pop();
        ll n = p.second;
        ll c = p.first;
        if(d[n] >= 0) continue;
        d[n] = c;
        for(auto &e : G[n]){
            if(d[e.first] >= 0) continue;
            pq.push({d[n]+e.second,e.first});
        }
    }
    cout << d[N-1] << endl;
    return 0;
}