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;

long long N, M;
vector<pair<long long, long long> > tree[100005];
long long ans[100005];

int main(){
	cin >> N >> M;

	for(long long i = 0; i < 100005; i++){
		ans[i] = -1;
	}

	for(long long i = 0; i < M; i++){
		long long l, r, c;
		cin >> l >> r >> c; l--; r--;
		tree[l].push_back(make_pair(r, c));
	}

	//                    cost   , point
	priority_queue<pair<long long, long long>, vector<pair<long long, long long> >, greater<pair<long long, long long> > > Q;
	Q.push(make_pair(0, 0));

	while(!Q.empty()){
		long long cost = Q.top().first;
		long long point = Q.top().second;
		Q.pop();
		if(ans[point] >= 0){
			continue;
		}
		ans[point] = cost;

		for(long long i = 0; i < tree[point].size(); i++){
			long long nextcost = cost + tree[point][i].second;
			long long nextpoint = tree[point][i].first;
			if(ans[nextpoint] < 0){
				Q.push(make_pair(nextcost, nextpoint));
			}
		}

		if(point > 0 && ans[point - 1] < 0){
			Q.push(make_pair(cost, point - 1));
		}
	}

	cout << ans[N-1] << endl;
}