#include <bits/stdc++.h>
using namespace std;
 
int main() {
  long n,m;
  cin >> n >> m;
  vector<long> l(m),r(m),c(m),saitan(n,1000000000000000);
  vector<vector<long>> lr(n,vector<long>(0)),lc(n,vector<long>(0)),rc(n,vector<long>(0));
  multiset<long> ms;
  saitan[0]=0;
  for(long i=0;i<m;i++) {
    cin >> l[i] >> r[i] >> c[i];
    lr[l[i]-1].push_back(r[i]-1);
    lc[l[i]-1].push_back(c[i]);
  }
  for(long i=0;i<lr[0].size();i++) {
    ms.insert(lc[0][i]);
    rc[lr[0][i]].push_back(lc[0][i]);
  }
  for(long i=1;i<n;i++) {
    if(ms.size()!=0) saitan[i]=min(1000000000000000L,*(ms.begin()));
    for(long j=0;j<rc[i].size();j++) {
      ms.erase(ms.find(rc[i][j]));
    }
    for(long j=0;j<lr[i].size();j++) {
      ms.insert(lc[i][j]+saitan[i]);
      rc[lr[i][j]].push_back(lc[i][j]+saitan[i]);
    }
  }
  if(saitan[n-1]!=1000000000000000)cout << saitan[n-1] << endl;
  else cout << -1 << endl;
}
### Prompt

Your challenge is to write a cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
 
int main() {
  long n,m;
  cin >> n >> m;
  vector<long> l(m),r(m),c(m),saitan(n,1000000000000000);
  vector<vector<long>> lr(n,vector<long>(0)),lc(n,vector<long>(0)),rc(n,vector<long>(0));
  multiset<long> ms;
  saitan[0]=0;
  for(long i=0;i<m;i++) {
    cin >> l[i] >> r[i] >> c[i];
    lr[l[i]-1].push_back(r[i]-1);
    lc[l[i]-1].push_back(c[i]);
  }
  for(long i=0;i<lr[0].size();i++) {
    ms.insert(lc[0][i]);
    rc[lr[0][i]].push_back(lc[0][i]);
  }
  for(long i=1;i<n;i++) {
    if(ms.size()!=0) saitan[i]=min(1000000000000000L,*(ms.begin()));
    for(long j=0;j<rc[i].size();j++) {
      ms.erase(ms.find(rc[i][j]));
    }
    for(long j=0;j<lr[i].size();j++) {
      ms.insert(lc[i][j]+saitan[i]);
      rc[lr[i][j]].push_back(lc[i][j]+saitan[i]);
    }
  }
  if(saitan[n-1]!=1000000000000000)cout << saitan[n-1] << endl;
  else cout << -1 << endl;
}
```