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>

int ri() {
	int n;
	scanf("%d", &n);
	return n;
}

int main() {
	int n = ri();
	int m = ri();
	int l[m], r[m], c[m];
	for (int i = 0; i < m; i++) l[i] = ri() - 1, r[i] = ri() - 1, c[i] = ri();
	std::vector<int> hen[n];
	std::vector<int> finish[n + 1];
	int64_t cost_hen[m];
	for (int i = 0; i < m; i++) hen[l[i]].push_back(i), finish[r[i]].push_back(i);
	int64_t dp[n];
	for (int i = 0; i < m; i++) cost_hen[i] = l[i] ? 1000000000000000000 : c[i];
	for (int i = 0; i < n; i++) dp[i] = 1000000000000000000;
	dp[0] = 0;
	std::multiset<int64_t> costs;
	std::set<int> cur;
	for (int i = 0; i + 1 < n; i++) {
		for (auto j : hen[i]) cur.insert(j), costs.insert(c[j] + dp[i]);
		for (auto j : finish[i]) cur.erase(j), costs.erase(costs.find(c[j] + dp[l[j]]));
		if (costs.size()) dp[i + 1] = *costs.begin();
	}
	std::cout << (dp[n - 1] >= 1000000000000000000 ? -1 : dp[n - 1]) << std::endl;
	return 0;
}