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;

const int MAXN = 100005;

int N, M, Li, Ri, Ci;
vector < pair <int, int> > evs[MAXN];

int main(){
	scanf("%d %d", &N, &M);
	while(M--){
		scanf("%d %d %d", &Li, &Ri, &Ci);
		Li--, Ri--;
		
		evs[Li].push_back(make_pair(Ri, Ci));
	}
	
	set < pair <long long, int> > S;
	S.insert(make_pair(0, 0));
	
	for(int i = 0; i < N; i++){
		while(!S.empty() && S.begin()->second < i)
			S.erase(S.begin());
			
		if(S.empty()){
			printf("-1\n");
			return 0;
		}
		
		long long dist = S.begin()->first;
		
		if(i == N - 1){
			printf("%lld\n", dist);
			return 0;
		}
		
		for(pair <int, int> e : evs[i])
			S.insert(make_pair(dist + e.second, e.first));
	}
	
	return 0;
}