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 <vector>
#include <queue>
#include <algorithm>

using namespace std;

typedef long long ll;
typedef pair<ll,int> P;
const ll INF = 1e17;

vector<P> g[100005];

ll dist[100005];

int main() {
  int n,m;
  cin >> n >> m;

  for(int i = n - 1; i > 0; i--) {
    g[i].push_back(P(0,i-1));
  }
  

  for(int i = 0; i < m; i++) {
    int l,r,c;
    cin >> l >> r >> c;
    l--; r--;
    g[l].push_back(P(c,r));
  }

  for(int i = 0; i < n; i++) dist[i] = INF;

  priority_queue<P,vector<P>, greater<P> > q;

  dist[0] = 0;
  q.push(P(0,0));

  while(!q.empty()) {
    P v = q.top();
    q.pop();

    if(v.first > dist[v.second]) continue;

    for(int i = 0; i < g[v.second].size(); i++) {
      P u = g[v.second][i];
      if(dist[u.second] > dist[v.second] + u.first) {
	dist[u.second] = dist[v.second] + u.first;
	q.push(P(dist[u.second],u.second));
      }
    }
  }

  if(dist[n-1] == INF) cout << -1 << endl;
  else cout << dist[n-1] << endl;
}