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;
using pll = pair<long, long>;
#define rep(i,n) for(int i=0; i<(int)(n);i++)
#define MOD 1000000007
#define INF 1e18

int main(){
  int n,m;
  cin >> n >> m;
  vector<vector<pll>> Graph(n);
  rep(i,n-1){
    Graph.at(i+1).push_back(pll(i,0));
  }
  rep(i,m){
    long l,r,c;
    //cin >> l >> r >> c;
    scanf("%ld%ld%ld",&l,&r,&c);
    Graph.at(l-1).push_back(pll(r-1,c));
  }
  priority_queue<pll,vector<pll>, greater<pll>> que;
  vector<long> dis(n,INF);
  dis.at(0) = 0;
  que.push(pll(0,0));
  while(que.size()){
    pll P = que.top();
    que.pop();
    long v = P.second;
    long d = P.first;
    if(d > dis.at(v)) continue;
    for(pll nP : Graph.at(v)){
      long nv = nP.first;
      long cost = nP.second;
      if(d + cost < dis.at(nv)){
        dis.at(nv) = d + cost;
        que.push(pll(dis.at(nv),nv));
      }
    }
  }
  //rep(i,n-1) cout << dis.at(i) << endl;
  if(dis.at(n-1) == INF) cout << -1 << endl;
  else cout << dis.at(n-1) << endl;
}