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 <iostream>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;
using ll = long long;
constexpr ll INF = 1LL << 60;

ll N, M, L, R, C;
vector<pair<ll, ll>> g[100010];

int main() {
  cin >> N >> M;
  for(int i = 0; i < M; ++i) {
    cin >> L >> R >> C;
    --L, --R;
    g[L].emplace_back(R, C);
  }
  for(int i = 1; i < N; ++i) 
    g[i].emplace_back(i-1, 0);

  vector<ll> dist(N, INF);
  priority_queue<pair<ll, ll>, vector<pair<ll, ll>>, greater<pair<ll, ll>>> pq;
  dist[0] = 0;
  pq.emplace(0, 0);
  while(pq.size()) {
    ll d = pq.top().first;
    ll v = pq.top().second;
    pq.pop();
    for(auto next : g[v]) {
      ll nv = next.first;
      ll nc = next.second;
      if(dist[nv] > d + nc) {
        dist[nv] = d + nc;
        pq.emplace(dist[nv], nv);
      }
    }
  }
  if(dist[N-1] == INF) cout << -1;
  else cout << dist[N-1];
  return 0;
}