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>
#define INF 2000000000000000000
#define ll long long
using namespace std;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  ll N, M;
  cin >> N >> M;
  vector<vector<vector<ll>>> path(N);
  for (ll i = 0; i < M; ++i) {
    ll l, r, c;
    cin >> l >> r >> c;
    path.at(l - 1).push_back({r - 1, c});
  }
  multiset<ll> st;//min
  vector<vector<ll>> imos(N + 1);
  st.insert(0);
  imos.at(1).push_back(0);
  for (ll i = 0; i < N; ++i) {
    for (ll j = 0; j < imos.at(i).size(); ++j) {
      st.erase(st.lower_bound(imos.at(i).at(j)));
    }
    if (st.size() == 0) {
      cout << "-1" << "\n";
      return 0;
    }
    ll m = *st.begin();
    for (ll j = 0; j < path.at(i).size(); ++j) {
      ll cost = m + path.at(i).at(j).at(1);
      st.insert(cost);
      imos.at(path.at(i).at(j).at(0) + 1).push_back(cost);
    }
  }
  // for (ll i = 0; i < N + 1; ++i) {
  //   cout << "i" << i << "\n";
  //   for (ll j = 0; j < imos.at(i).size(); ++j) {
  //     cout << imos.at(i).at(j) << " ";
  //   }
  //   cout << "\n";
  // }
  cout << *st.begin() << "\n";
}