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 vector<ll> vec;
typedef vector<vec> mat;
typedef pair<ll,ll> pll;
const ll mod=1e9+7;
//const ll mod=998244353;
const ll inf=5e15;

int main() {
  ll n,m;
  cin >> n >> m;
  vector<map<ll,ll>> E(n);
  for(ll i=1;i<n;i++) E[i][i-1]=0;
  for(ll i=0;i<m;i++) {
    ll l,r,c;
    cin >> l >> r >> c;
    l--;r--;
    if(E[l][r]) {
      E[l][r]=min(E[l][r],c);
    }
    else {
      E[l][r]=c;
    }
  }
  priority_queue<pll,vector<pll>,greater<pll>> q;
  vec d(n,inf);
  q.push(pll(0,0));
  while(q.size()) {
    ll dis=q.top().first;
    ll f=q.top().second;
    q.pop();
    if(d[f]<dis) continue;
    d[f]=dis;
    for(auto p:E[f]) {
      ll ndis=dis+p.second;
      ll t=p.first;
      if(d[t]>ndis) {
        q.push(pll(ndis,t));
      }
    }
  }
  if(d[n-1]==inf) d[n-1]=-1;
  cout << d[n-1] << endl;
}