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>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pii;
const ll maxn = 1e6+10;
const ll inf = 1e14;

vector<pii> ve[maxn];
ll dis[maxn];
priority_queue<pii,vector<pii>,greater<pii> > heap;
int N,M;

void init(){
	for(int i = 1;i<=N;i++) dis[i] = inf;
	for(int i = 1;i<=N;i++) ve[i].push_back({0,i-1});
	
}

ll Dji(){
	heap.push({0,1});
	dis[1] = 0;
	
	while(heap.size()){
		auto cur = heap.top();heap.pop();
		ll d = cur.first,id = cur.second;
		for(auto v:ve[id]){
			ll w = v.first,id2 = v.second;
			if(d+w<dis[id2]){
				dis[id2] = d+w;
				heap.push({dis[id2],id2});
			}
		}
	}
	if(dis[N] == inf) return -1;
	return dis[N];
}

int main(){
	cin>>N>>M;
	init();
	int a,b,c;
	for(int i = 0;i<M;i++){
		scanf("%d%d%d",&a,&b,&c);
		ve[a].push_back({c,b});
	}
	cout<<Dji()<<endl;
	return 0;
}