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 <iostream>
#include <vector>
#include <algorithm>
#include <climits>
using namespace std;
using uint = unsigned int;
using ull = unsigned long long;
using ll = long long;
 
int main() {
	
	uint N, M;
	cin >> N >> M;
	
	vector<pair<pair<uint, uint>, uint>> LRC(M);
	uint l, r, c;
	for(uint i=0; i<M; i++) {
		cin >> l >> r >> c;
		auto p = make_pair(l - 1, r - 1);
		LRC[i] = make_pair(p, c);
	}
	
	sort(LRC.begin(), LRC.end());
	
	vector<ll> d(N, -1);
	d[0] = 0;
	
	l = 0;
	r = 0;
	for(uint i=0; i<M; i++) {
		auto LR = LRC[i].first;
		l = LR.first;
		r = LR.second;
		c = LRC[i].second;
		
		if(d[l] == -1) {
			cout << -1;
			return 0;
		}
		
		ll dj = d[l] + c;
		for(uint j=r; j>l; j--) {
			if(d[j] == -1) {
				d[j] = dj;
			} else if (d[j] <= dj) {
				break;
			} else {
				d[j] = dj;
			}
		}
	}
	
	cout << d[N-1];
	
	return 0;
}