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<iostream>
#include<vector>
#include<algorithm>
#include<queue>
#define MAX_N 200000
#define INF 100000000000000
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
struct edge{
  ll to;
  ll cost;
};
vector<edge> G[MAX_N];
ll d[MAX_N];
signed main(){
  ll n,m,l,r,c;
  priority_queue<P,vector<P>,greater<P>> que;
  cin >> n >> m;
  for(int i=0;i<m;i++){
    cin >> l >> r >> c;
    l--;
    r--;
    G[l].push_back({r,c});
    G[r].push_back({l,c});
  }
  for(int i=1;i<n;i++){
    G[i].push_back({i-1,0});
    d[i] = INF;
  }
  d[0]=INF;
  que.push(P(0,0));
  while(!que.empty()){
    P tmp = que.top();
    que.pop();
    ll v,dis;
    v = tmp.second;
    dis = tmp.first;
    if(d[v]<=dis){
      continue;
    }
    d[v] = dis;
    for(int i=0;i<G[v].size();i++){
      que.push({dis+G[v][i].cost,G[v][i].to});
    }
  }
  if(d[n-1]==INF) d[n-1]=-1;
  cout << d[n-1]<<endl;
}