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 REP(i,n) for(int i=0;i<(n);++i)
using namespace std;
typedef long long ll;

const ll INF = 1LL << 60;

vector< vector< pair<int,int> > > to(100005);
ll dist[100005];

int main() {
  int N,M;
  cin >> N >> M;
  REP(i,M) {
    int l,r,c;
    cin >> l >> r >> c;
    l--; r--;
    to[l].emplace_back(r,c);
  }
  REP(i,N-1)
    to[i+1].emplace_back(i,0);

  REP(i,N)
    dist[i] =INF;
  dist[0] = 0;
  // ダイクストラ
  priority_queue< pair<ll, int>, vector< pair<ll,int> >, greater<pair<ll,int> > > q;
  q.push( make_pair(0, 0) );
  while( !q.empty() ) {
    auto v = q.top(); q.pop();
    if ( v.first > dist[v.second] ) continue;
    for ( auto nv: to[v.second] ) {
      if ( dist[nv.first] > dist[v.second] + nv.second ) {
        dist[nv.first] = dist[v.second] + nv.second;
        q.push( make_pair(dist[nv.first], nv.first) );
      }
    }
  }
  ll ans = (dist[N-1] == INF) ? -1 : dist[N-1];
  cout << ans << endl;
}