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>
#define rep(i,n) for(int i = 0; i < n; i++)
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
const int MAX_N = 100005;
const ll INF = 1e15;

struct edge
{
	int to,cost;
};

vector<edge> G[MAX_N];
ll d[MAX_N];

void init()
{
	rep(i,MAX_N)
	{
		d[i] = INF;
	}
}

int main()
{
	init();
	int N,M;
	edge e;
	cin >> N >> M;
	rep(i,N-1)
	{
		e.to = i;
		e.cost = 0;
		G[i+1].push_back(e);
	}
	rep(i,M)
	{
		int l,r,c;
		cin >> l >> r >> c;
		e.to = r-1;
		e.cost = c;
		G[l-1].push_back(e);
	}

	priority_queue<P, vector<P>, greater<P> > Q;
	d[0] = 0;
	Q.push(P(0,0));
	while(!Q.empty())
	{
		P p = Q.top();
		Q.pop();
		int v = p.second;
		if (d[v] < p.first) continue;
		for (int i = 0; i < G[v].size(); ++i)
		{
			edge e = G[v][i];
			if (d[e.to] > d[v] + e.cost)
			{
				d[e.to] = d[v] + e.cost;
				Q.push(P(d[e.to],e.to));
			}
		}
	}

	if (d[N-1] == INF) cout << "-1" << endl;
	else cout << d[N-1] << endl;
	return 0;
}