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;
#define rep(i,n) for(int i = 0; i < (n);i++)
#define sz(x) int(x.size())
typedef long long ll;
typedef pair<int,int> P;
using Graph = vector<vector<P>>;
const ll INF = 1e18;
int main(){
  int n, m;
  cin >> n >> m;
  Graph G(n);
  /*コスト０の逆辺 */
  for (int i = n-1; i > 0; i--) {
    G[i].push_back({0,i-1});
  }
  rep(i,m) {
    int l, r, c;
    cin >> l >> r >> c;
    l--; r--;
    G[l].push_back({c,r});
  }

  auto dijkstra = [&] {
    priority_queue<pair<ll,int>,vector<pair<ll,int>>, greater<pair<ll,int>>> que;
    vector<ll> d(n,INF);
    d[0] = 0;
    que.push({0,0});
    while (!que.empty()) {
      auto p = que.top();
      que.pop();
      if (d[p.second] < p.first) continue;
      for (auto v : G[p.second]) {
        if (d[v.second] <= d[p.second] + v.first) continue;
        que.push({d[p.second] + v.first, v.second});
        d[v.second] = d[p.second] + v.first;
      }
    }
    return d[n-1];
  };
  ll res = dijkstra();
  cout << (res == INF ? -1 : res) << endl;
  return 0;
}