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<string.h>
#include<queue>
using namespace std;
const long long N = 4e6 + 7;
struct node {
	long long to, cost;
}r;
struct que {
	long long x, y;
	bool friend operator < (que a, que b) {
		return a.x > b.x;
	}
}w;
long long vis[N];
long long dis[N];
vector<node>v[N];
long long n, m, x, y, z;
void push(long long x, long long y, long long z) {
	r.to = y; r.cost = z;
	v[x].push_back(r);
}
long long inf = 214722121483647;
priority_queue<que> q;
long long slove(long long x) {
	fill(dis, dis + N, inf);
	w.x = 0; w.y = 1;
	dis[1] = 0;
	q.push(w);
	while (!q.empty()) {
		w = q.top(); q.pop();
		long long pos = w.y;
		for (long long i = 0; i < v[pos].size(); i++) {
			node e = v[pos][i];
			if (dis[e.to] > dis[pos] + e.cost) {
				dis[e.to] = dis[pos] + e.cost;
				w.x = dis[e.to]; w.y = e.to;
				q.push(w);
			}
		}
		
	}
	return dis[n];
}
int main() {
	cin >> n >> m;
	for (long long i = 1; i <= n; i++) {
		push(i, i - 1, 0);
	}
	while (m--) {
		cin >> x >> y >> z;
		push(x, y, z);
	}
	if (slove(1) == inf) cout << "-1\n";
	else cout << dis[n] << '\n';
	return 0;
}