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;

const int N=100000;

int n,m;
struct seg{
  int l,r,v;
}a[N+9];
vector<int>vl[N+9],vr[N+9];

void into(){
  scanf("%d%d",&n,&m);
  for (int i=1;i<=m;++i){
	scanf("%d%d%d",&a[i].l,&a[i].r,&a[i].v);
	vl[a[i].l].push_back(i);
	vr[a[i].r].push_back(i);
  }
}

multiset<LL>s;
LL dp[N+9];

void Get_dp(){
  for (int i=1;i<=n;++i){
	if (i>1&&s.empty()) {dp[n]=-1;return;}
	dp[i]=i==1?0:*s.begin();
	for (int vs=vr[i].size(),j=0;j<vs;++j) s.erase(s.find(dp[a[vr[i][j]].l]+a[vr[i][j]].v));
	for (int vs=vl[i].size(),j=0;j<vs;++j) s.insert(dp[a[vl[i][j]].l]+a[vl[i][j]].v);
  }
}

void work(){
  Get_dp();
}

void outo(){
  printf("%lld\n",dp[n]);
}

int main(){
  into();
  work();
  outo();
  return 0;
}