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 pb push_back
#define all(v) (v).begin() , (v).end()
#define popcnt(x) __builtin_popcount(x)
#define inf 0x3f3f3f3f
#define watch(x) cout << (#x) << " is " << (x) << endl
#define rand() (rand() << 15 | rand())
using namespace std;
typedef long long  ll;
#define EPS 1e-9
#define PI acos(-1.0)
const int N = 1e5 + 4;

vector<pair<int , int>> add[N];
vector<long long> er[N];
long long dp[N];
int main() {
   ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
   int n , m;
   cin >> n >> m;
   for(int i = 0 ; i < m ; ++i){
      int l , r , c;
      cin >> l >> r >> c;
      add[l].push_back({r , c});
   }


   multiset<long long> ms;
   for(int i = 1 ; i <= n ; ++i){
      for(long long c : er[i])
         ms.erase(ms.find(c));
      if(i > 1)dp[i] = (ms.empty() ? -LLONG_MAX / 2 : *ms.begin());
      else dp[i] = 0;
      for(auto p : add[i]){
         ms.insert(dp[i] + p.second);
         er[p.first + 1].push_back(dp[i] + p.second);
      }
   }

   cout << (dp[n] < 0 ? -1 : dp[n]) << "\n";
}