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;
typedef bool boolean;

#define ll long long

const ll llf = (signed ll) (~0ull >> 3);

const int N = 1e5 + 5;

typedef class Edge {
	public:
		int L, R;
		ll c;

		void read() {
			scanf("%d%d%lld", &L, &R, &c);
		}
		boolean operator < (Edge b) const {
			return c > b.c;
		}
} Edge;

int n, m;
ll f[N];
vector<Edge> G[N];
priority_queue<Edge> Q;

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++)
		f[i] = llf;
	f[1] = 0;
	for (int i = 1; i <= m; i++) {
		Edge tmp;
		tmp.read();
		G[tmp.L].push_back(tmp);
	}
	for (int i = 1; i <= n; i++) {
		while (!Q.empty() && Q.top().R < i)
			Q.pop();
		if (i > 1) {
			if (Q.empty()) {
				puts("-1");
				return 0;
			}
			f[i] = Q.top().c;
		}
		for (auto e : G[i]) {
			e.c += f[i];
			Q.push(e);
		}
	}
	printf("%lld\n", f[n]);
	return 0;
}