#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;

struct edge{int to; ll cost;};
typedef pair<ll, int> P;
const ll INF = 1e18;
const int MAX_V = 1e5;

vector<edge> G[MAX_V];
vector<ll> d(MAX_V,INF); 

void dijkstra(int s){
  priority_queue<P, vector<P>, greater<P>> que;
  d[s] = 0;
  que.push(P(0, s));
  while(!que.empty()){
    P p = que.top();
    que.pop();
    int v = p.second;
    if(d[v] < p.first) continue;
    rep(i, G[v].size()){
      edge e = G[v][i];
      if(d[e.to] > d[v] + e.cost){
        d[e.to] = d[v] + e.cost;
        que.push(P(d[e.to], e.to));
      }
    }
  }
}

int main(){
  int n, m;
  cin >> n >> m;
  for(int i = n-1; i >= 1; i--) G[i].push_back(edge{i-1,0});
  rep(i,m) {
    int l, r; ll c;
    cin >> l >> r >> c;
    l--; r--;
    G[l].push_back(edge{r,c});
  }
  dijkstra(0);
  if(d[n-1] == INF) cout << -1 << endl;
  else cout << d[n-1] << endl;
  return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;

struct edge{int to; ll cost;};
typedef pair<ll, int> P;
const ll INF = 1e18;
const int MAX_V = 1e5;

vector<edge> G[MAX_V];
vector<ll> d(MAX_V,INF); 

void dijkstra(int s){
  priority_queue<P, vector<P>, greater<P>> que;
  d[s] = 0;
  que.push(P(0, s));
  while(!que.empty()){
    P p = que.top();
    que.pop();
    int v = p.second;
    if(d[v] < p.first) continue;
    rep(i, G[v].size()){
      edge e = G[v][i];
      if(d[e.to] > d[v] + e.cost){
        d[e.to] = d[v] + e.cost;
        que.push(P(d[e.to], e.to));
      }
    }
  }
}

int main(){
  int n, m;
  cin >> n >> m;
  for(int i = n-1; i >= 1; i--) G[i].push_back(edge{i-1,0});
  rep(i,m) {
    int l, r; ll c;
    cin >> l >> r >> c;
    l--; r--;
    G[l].push_back(edge{r,c});
  }
  dijkstra(0);
  if(d[n-1] == INF) cout << -1 << endl;
  else cout << d[n-1] << endl;
  return 0;
}
```