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 <queue>
#include <functional>
using namespace std;

long long N, M, L[1 << 18], R[1 << 18], C[1 << 18];
long long dist[1 << 18];
vector<pair<int, int>> X[1 << 18];
priority_queue<pair<long long, int>, vector<pair<long long, int>>, greater<pair<long long, int>>> Q;

int main() {
	cin >> N >> M;
	for (int i = 1; i <= M; i++) cin >> L[i] >> R[i] >> C[i];
	for (int i = 1; i <= M; i++) X[L[i]].push_back(make_pair(R[i], C[i]));
	for (int i = N - 1; i >= 1; i--) X[i + 1].push_back(make_pair(i, 0));
	for (int i = 1; i <= N; i++) dist[i] = (1LL << 60);

	Q.push(make_pair(0, 1)); dist[1] = 0;
	while (!Q.empty()) {
		int pos = Q.top().second; Q.pop();
		for (pair<int, int> i : X[pos]) {
			int to = i.first;
			long long cost = i.second;
			if (dist[to] > dist[pos] + cost) {
				dist[to] = dist[pos] + cost;
				Q.push(make_pair(dist[to], to));
			}
		}
	}

	if (dist[N] == (1LL << 60)) cout << "-1" << endl;
	else cout << dist[N] << endl;
	return 0;
}