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 <cstdio>
#include <vector>
#include <queue>
#include <utility>
#include <functional>

struct edge
{
  int to;
  long long cost;
};
std::vector<edge> edges[100000];
  
int main()
{
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i < n; ++i)
  {
    edges[i].emplace_back(edge{i - 1, 0});
  }
  for (int i = 0; i < m; ++i)
  {
    int a, b;
    long long c;
    scanf("%d%d%lld", &a, &b, &c);
    --a, --b;
    edges[a].emplace_back(edge{b, c}); 
  }
  long long dist[100000];
  using pair = std::pair<long long, int>;
  std::fill(dist, dist + n, 1e18);
  dist[0] = 0;
  std::priority_queue<pair, std::vector<pair>, std::greater<pair>> q;
  q.emplace(0, 0);
  while (q.size() > 0)
  {
    auto p = q.top(); q.pop();
    if (p.first > dist[p.second]) continue;
    for (const auto& e : edges[p.second])
    {
      long long next = p.first + e.cost;
      if (dist[e.to] > next)
      {
        dist[e.to] = next;
        q.emplace(next, e.to);
      }
    }
  }
  printf("%lld\n", dist[n - 1] == 1e18 ? -1 : dist[n - 1]);
  return 0;
}