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 = 1e5+100;
const ll inf = 0x3f3f3f3f3f3f3f3fLL;
#define fr(i,a,b) for(int i = a;i <= b; ++i)
#define nfr(i,a,b) for(int i = a;i >= b; --i)

struct bit{
   ll val[N];
   void init(){
      memset(val,0x3f,sizeof val);
   }
   void upd(int x,ll v){
      for(;x < N;x += (x&(-x)))val[x] = min(val[x],v);
   }
   inline ll query(int x){
      ll res = 0x3f3f3f3f3f3f3f3fLL;
      for(;x;x -= (x&(-x)))res = min(res,val[x]);
      return res;
   }
}s;

struct query{
   int l,r,c;
   inline bool operator < (const query &b) const {
      return r < b.r;
   }
}q[N];

int n,m;
ll f[N];
int main(){
   scanf("%d%d",&n,&m);
   fr(i,1,m){
      scanf("%d%d%d",&q[i].l,&q[i].r,&q[i].c);
   }
   sort(q+1,q+m+1);

   s.init();
   s.upd(n,0);
   fr(i,1,m){
      f[i] = q[i].c+s.query(n+1-q[i].l);
      s.upd(n+1-q[i].r,f[i]);
   }
   ll w = s.query(1);
   if(w == inf)puts("-1");
   else printf("%lld\n",w);
   return 0;
}