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;
typedef long long ll;

int main() {
  int N, M;
  cin >> N >> M;
  vector<vector<pair<ll, int>>> cost(N);
  for (int i = 1; i < N; i++) {
    cost[i].push_back(make_pair(0, i - 1));
  }
  for (int i = 0; i < M; i++) {
    int L, R, C;
    cin >> L >> R >> C;
    L--, R--;
    cost[L].push_back(make_pair(C, R));
  }

  vector<ll> to(N);
  for (int i = 1; i < N; i++) to[i] = 1e15;
  priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>>
      que;
  que.push(make_pair(0, 0));
  while (!que.empty()) {
    int c = que.top().first, now = que.top().second;
    que.pop();
    if (c > to[now]) continue;
    for (int i = 0; i < cost[now].size(); i++) {
      if (to[cost[now][i].second] > to[now] + cost[now][i].first) {
        to[cost[now][i].second] = to[now] + cost[now][i].first;
        que.push(make_pair(to[cost[now][i].second], cost[now][i].second));
      }
    }
  }
  cout << (to[N - 1] == 1e15 ? -1 : to[N - 1]) << endl;
}