#include <bits/stdc++.h>
using namespace std;
using Int = int_fast64_t;

using P = pair<Int, Int>;
using PP = pair<P, Int>;
Int n, m;
vector<PP> v;
vector<Int> d;

Int getmin(Int p){
	Int ret = 1e18;
	for(Int i=p; i>0; i-=i&-i) ret = min(ret, d[i]);
	return ret;
}
void update(Int p, Int q){
	for(Int i=p; i<=n; i+=i&-i) d[i] = min(d[i], q);
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin >> n >> m;
	v.resize(m);
	for(Int i=0; i<m; ++i){
		Int l, r, c; cin >> l >> r >> c;
		v[i] = PP(P(l, r), c);
	}
	sort(v.rbegin(), v.rend());
	d.resize(n+1, 1e18);
	d[n] = 0;
	for(Int i=0; i<m; ++i){
		Int l = v[i].first.first;
		Int r = v[i].first.second;
		Int c = v[i].second;
		Int dr = getmin(r);
		update(l, dr+c);
	}
	if(d[1] == 1e18) cout << -1 << "\n";
	else cout << d[1] << "\n";
}
### 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;
using Int = int_fast64_t;

using P = pair<Int, Int>;
using PP = pair<P, Int>;
Int n, m;
vector<PP> v;
vector<Int> d;

Int getmin(Int p){
	Int ret = 1e18;
	for(Int i=p; i>0; i-=i&-i) ret = min(ret, d[i]);
	return ret;
}
void update(Int p, Int q){
	for(Int i=p; i<=n; i+=i&-i) d[i] = min(d[i], q);
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin >> n >> m;
	v.resize(m);
	for(Int i=0; i<m; ++i){
		Int l, r, c; cin >> l >> r >> c;
		v[i] = PP(P(l, r), c);
	}
	sort(v.rbegin(), v.rend());
	d.resize(n+1, 1e18);
	d[n] = 0;
	for(Int i=0; i<m; ++i){
		Int l = v[i].first.first;
		Int r = v[i].first.second;
		Int c = v[i].second;
		Int dr = getmin(r);
		update(l, dr+c);
	}
	if(d[1] == 1e18) cout << -1 << "\n";
	else cout << d[1] << "\n";
}
```