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>
#include <queue>
using namespace std;
typedef pair<long long, int> P; 
int main(){
	int N, M; cin >> N >> M; 
	long long INF = 1e18; 
	vector<P> E[100010] = {};  
	for(int i = 1; i < N; i++){
		E[i].push_back(make_pair(0LL, i-1)); 
	}
	for(int i = 0; i < M; i++){
		int L, R; long long C; cin >> L >> R >> C; 
		L--; R--; 
		E[L].push_back(make_pair(C, R)); 
	}
	long long dist[100010] = {}; 
	fill(dist+1, dist+N, INF); 
	priority_queue<P, vector<P>, greater<P>> que;
	que.push(make_pair(0LL, 0)); 
	while(que.size()){
		P p = que.top();
		que.pop(); 
		int now = p.second; 
		for(int i = 0; i < E[now].size(); i++){
			if(dist[now] + E[now].at(i).first < dist[E[now].at(i).second]){
				dist[E[now].at(i).second] = dist[now] + E[now].at(i).first;
				que.push(make_pair(dist[now] + E[now].at(i).first, E[now].at(i).second)); 
			}
		}
	}
	if(dist[N-1] == INF) cout << -1 << endl; 
	else cout << dist[N-1] << endl; 
}