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>
typedef long long LL;
#define FOR(i,a,b) for(LL i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)

using namespace std;
LL p = 1000000007ll;

int main(void)
{
  LL n,m;
  cin >> n >> m;
  vector<int> l(m),r(m),c(m);
  REP(i,m) cin >> l[i] >> r[i] >> c[i];
  vector<vector<pair<int,int>>> nei(n+1);
  REP(i,m) {
    nei[l[i]].emplace_back(r[i], c[i]);
  }
  FOR(i,1,n) {
    nei[i+1].emplace_back(i,0);
  }

  priority_queue<pair<LL,LL>> q;
  set<LL> already;
  // negc, pos
  q.emplace(0,1);
  while(!q.empty()) {
    LL negc,pos;
    tie(negc,pos) = q.top();
    q.pop();
    if(!already.insert(pos).second) continue;
    if(pos == n) {
      cout << -negc << endl;
      return 0;
    }
    for(auto pa:nei[pos]) {
      q.emplace(negc - pa.second, pa.first);
    }
  }
  cout << -1 << endl;
  return 0;
}