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;
typedef long long ll;
#define x first
#define y second
#define mp make_pair
#define pb push_back
#define sz(a) (int)(a.size())
const int mod = 998244353;
const int MAX = 100005;
vector<pair<int , int> > lc[MAX];
ll dp[MAX];
vector<pair<int , ll> > DP;
int main()
{	
	int n , m;cin >> n >> m;
	for(int i = 0; i < m; i++)
	{
		int l , r , c; cin >> l >> r >> c;
	
		l--;r--;
		lc[r].pb({l , c});
	}
	dp[0] = 0;
	DP.pb({0 , 0});
	for(int i = 1; i < n; i++)
	{
		dp[i] = mod * (ll) mod;
		for(auto x : lc[i])
		{
			//cout << x.x << " " << x.y << " " << i << endl;
			dp[i] = min(dp[i] , x.y + lower_bound(DP.begin() , DP.end() , mp(x.x , (ll)-mod))->y);
		}
		//cout << dp[i] << endl;
		while(sz(DP) &&DP.back().y >= dp[i])
			DP.pop_back();
		DP.pb({i , dp[i]});
	}
	cout << ((dp[n - 1] == mod * (ll)mod) ? -1 : dp[n - 1]);
		
}