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;

int main() {
  const long inf = 1l << 62;
  int n, m;
  cin >> n >> m;
  vector<long> d(n + 1, inf);
  d.at(1) = 0;
  vector<vector<pair<int, long>>> to(n + 1);
  for (int i = 0; i < m; i++) {
    int l, r, c;
    cin >> l >> r >> c;
    to.at(l).emplace_back(r, c);
  }
  for (int i = 2; i < n; i++) {
    to.at(i).emplace_back(i - 1, 0);
  }
  priority_queue<pair<long, int>> q;
  q.emplace(0, 1);
  while (!q.empty()) {
    auto t = q.top();
    q.pop();
    int current = t.second;
    for (auto next : to.at(current)) {
      if (d.at(current) + next.second < d.at(next.first)) {
        d.at(next.first) = d.at(current) + next.second;
        q.emplace(-d.at(next.first), next.first);
      }
    }
  }
  if (d.at(n) == inf) {
    cout << -1 << endl;
    return 0;
  }
  cout << d.at(n) << endl;
}