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>

#define rep(i,n) for(int i = 0; i < (n); i++)
#define rrep(i,n) for(int i = (n)-1; i >= 0; i--)
#define rep1(i,n) for(int i = 1; i <= (n); i++)
#define rrep1(i,n) for(int i = (n); i > 0; i--)

#define ll long long
#define pi pair<int, int>
#define pll pair<ll, ll>

#define MOD 1000000007
#define INF 1000000000000000LL

using namespace std;

int main(){
  int n, m; cin>>n>>m;
  vector<vector<pll>> g(n, vector<pll>());
  rrep1(i, n-1)g[i].push_back({0, i-1});
  rep(i, m){
    ll l, r, c; cin>>l>>r>>c; l--;r--;
    g[l].push_back( {c, r} );
  }

  priority_queue<pll, vector<pll>, greater<pll>> pq;
  vector<ll> d(n);
  rep(i, n)d[i]=INF;
  d[0]=0;
  pq.push( {0, 0} );
  while(pq.size()){
    auto p = pq.top();pq.pop();
    if(p.first > d[p.second])continue;
    d[p.second] = p.first;
    for(auto np: g[p.second]){
      if(d[np.second] <= np.first+p.first)continue;
      d[np.second] = np.first+p.first;
      pq.push( {d[np.second], np.second} );
    }
  }

  cout<<(d[n-1] == INF ? -1 : d[n-1])<<endl;

  return 0;
}