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<algorithm>
#include<set>
#include<vector>
#define int long long
#define mod 1000000007
#define for0(i, n) for(int i = 0; i < (n); i++)
#define for1(i, n) for(int i = 1; i <= (n);i++)
using namespace std;
int n, m;
vector<pair<pair<int, int>, int>>v;
set<pair<int, pair<int, int>>>s;
int a[123456], p = 1;
signed main() {
	a[1] = 0;
	cin >> n >> m;
	for0(i, m) {
		pair<pair<int, int>, int>p1;
		cin >> p1.first.first >> p1.first.second >> p1.second;
		v.push_back(p1);
	}
	sort(v.begin(), v.end());
	int q = 0;
	while (p < n) {
		int q1 = q;
		while (v[q].first.first <= p && q < m) {
			v[q].second += a[v[q].first.first];
			s.insert(make_pair(v[q].second, v[q].first));
			q++;
		}
		if (s.size() == 0 && q1 == q) { cout << "-1\n"; return 0; }
		pair<int, pair<int, int>> p1 = *s.begin(); s.erase(s.begin());
		for (int j = p + 1; j <= p1.second.second; j++)a[j] = p1.first;
		p = p1.second.second;
	}
	cout << a[n] << endl;
}