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;
typedef pair<ll,ll> pii;

#define fi first
#define se second
#define mp make_pair
#define fastIO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int N = (int)1e5 + 9;
vector<pii> L[N];

int main(){
  fastIO;
  int n, m;
  cin >> n >> m;
  int l, r, c;
  for(int i = 1; i <= m; i ++ ){
    cin >> l >> r >> c;
    L[l].push_back(mp(r, c));
  }
  priority_queue<pii, vector<pii>, greater<pii>> ff;
  ff.push(mp(0ll, 1));
  ll mx;
  for(int i = 1; i <= n; i ++ ){
    while(!ff.empty() && i > ff.top().se){
      ff.pop();
    }
    if(ff.empty()){
      cout << "-1\n";
      return 0;
    }
    mx = ff.top().fi;
    if(i == n){
      cout << mx << "\n";
      return 0;
    }
    for(auto x : L[i]){
      ff.push(mp(mx + x.se, x.fi));
    }
  }
  return 0;
}