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;
const long long oo = 1LL << 60;

int n;
vector<pair<int, int>> a[100100];
vector<pair<long long, int>> out[100100];

bool cmp(pair<int, int> u, pair<int, int> v)
{
  if (u.first != v.first)
    return u.first < v.first;
  return u.second > v.second;
}

int main()
{
  ios_base::sync_with_stdio(0);
  cin.tie(0);

  int m;
  cin >> n >> m;
  while (m--)
  {
    int l, r, c;
    cin >> l >> r >> c;
    a[l].push_back({c, r});
  }

  set<pair<long long, int>> s;
  for (int i = 1; i <= n; i++)
  {
    long long dist = oo;
    if (i == 1) dist = 0;
    else if (s.empty()) dist = oo;
    else dist = s.begin()->first;

    if (i == n)
    {
      cout << (dist == oo ? -1 : dist) << endl;
      return 0;
    }

    if (dist != oo)
    {
      sort(a[i].begin(), a[i].end(), cmp);
      int r = i;
      for (auto u : a[i])
        if (u.second > r)
        {
          s.insert({dist + u.first, i});
          out[u.second].push_back({dist + u.first, i});
          r = u.second;
        }
    }

    for (auto u : out[i])
      s.erase(u);
  }
}