#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<ll, int>;
#define rep(i,n) for(int i=0; i<(n); i++)
const int MX = 100005;
int n, m, l, r, c;
ll dis;
vector<P> a[MX];
priority_queue<P, vector<P>, greater<P>> qu;

int main(){
	scanf("%d%d", &n, &m);
	rep(i,m){
		scanf("%d%d%d", &l, &r, &c);
		l--; r--;
		a[l].push_back(P(c,r));
	}
	rep(i,n){
		while(!qu.empty() && qu.top().second < i) qu.pop();
		if(i && qu.empty()) return 0*puts("-1");
		dis = i ? qu.top().first : 0;
		for(P p: a[i]) qu.push(P(p.first+dis,p.second));
	}
	printf("%lld\n", dis);
}
### Prompt

Please create a solution in CPP 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;
using ll = long long;
using P = pair<ll, int>;
#define rep(i,n) for(int i=0; i<(n); i++)
const int MX = 100005;
int n, m, l, r, c;
ll dis;
vector<P> a[MX];
priority_queue<P, vector<P>, greater<P>> qu;

int main(){
	scanf("%d%d", &n, &m);
	rep(i,m){
		scanf("%d%d%d", &l, &r, &c);
		l--; r--;
		a[l].push_back(P(c,r));
	}
	rep(i,n){
		while(!qu.empty() && qu.top().second < i) qu.pop();
		if(i && qu.empty()) return 0*puts("-1");
		dis = i ? qu.top().first : 0;
		for(P p: a[i]) qu.push(P(p.first+dis,p.second));
	}
	printf("%lld\n", dis);
}
```