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;
#define int long long
const int MOD = 1000000007;
 pair<pair<int,int>,int> p[100001];
signed main() {
 int n,m;
 cin>>n>>m;

 int c,l,r;
 for(int i=0;i<m;i++){
   cin>>l>>r>>c;
   p[i]=make_pair(make_pair(l-1,r-1),c);
 }
 int dp[n];sort(p,p+m);fill(dp,dp+n,MOD*MOD);
 dp[0]=0;int j=0;
  priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>> que;
   while(j<m&&p[j].first.first==0){
     que.push(make_pair(dp[0]+p[j].second,p[j].first.second));j++;
   }

 for(int i=1;i<n;i++){
   while(!que.empty()&&que.top().second<i)que.pop();
   if(!que.empty())dp[i]=que.top().first;
   while(j<m&&p[j].first.first<i)j++;
   while(j<m&&p[j].first.first==i){
     que.push(make_pair(dp[i]+p[j].second,p[j].first.second));j++;
   }
 }
 if(dp[n-1]>=MOD*MOD)cout<<-1;
 else cout<<dp[n-1];






return 0;
}